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56

I think that most of the answers here are incorrect since it has nothing to do with decreasing resistance of rubber. In fact, the force required to stretch the balloon increases, not decreases while inflating. It's similar to stretching a string, ie. the reaction force is proportional to the increase in length of the string - this is why there is a point ...


31

Take a strip of balloon rubber and pull it. It will get harder the more you pull. So why is it that inflating the balloon gets easier (at least long before the breaking point)? The balloon starts with very high curvature, so the air pressure is distorting each spot on its surface a lot relative to its 1cm neighbors for example. All the rubber's tension ...


17

When in doubt, use mathematics. Imagine the balloon as a sphere (close enough for this answer) of initial radius $r_0$ and thickness $t$. Let's inflate it just a little bit (to radius $r_0 + \Delta r$). Now we can take a look at what happens by taking a cut through the equator of the sphere. The total circumference at the equator is $2\pi r$; with the ...


15

Like Dev said above, the material your typical round balloon is made from has a non-linear stress strain curve. When just starting to inflate it is fairly stiff, but then as it starts to blow up the stiffness goes down somewhat until it approaches its maximum size. We measured this in my undergraduate advanced lab class, and while I don't have the data ...


15

In addition to the question of bend radius - there is also an effect of surface scratches. Most materials are very strong - they fail because a surface flaw allows a stress concentration - ie a crack to form. glass fibre has a very smooth surface because of the way it is made and can be put under high stress without cracking. You can show this with a thick ...


14

The volume of a balloon grows linear, while the surface (which you actually stretch) doesn't. So although you're blowing the same amount of air into a balloon, you don't stretch the surface as much as in the beginning.


11

What evidence do you have that there is a consistent theory of continuum mechanics? Certainly, when looked at through a macroscope, the universe looks like it behaves according to continuum mechanics, but this completely breaks down on the microscopic level. So you can't justify a consistent theory of continuum mechanics by using the universe. There's no ...


7

For me the best is "Landau, Lifschitz: Vol. 7"...


6

If you collide two ideal billiard balls, then that would be what you would call a perfectly elastic collision. If you have a large dense collection of billiard balls, and you slam a new one into the collection, then there are a whole lot of elastic collisions, transfering energy and momentum in many ways that you would be hard-pressed to calculate exactly. ...


6

Let us first summarize what do we actually experience when inflating the balloon. For the very first bit of volume, we have to exert a lot of energy. Or alternatively, we have to apply a lot of pressure coming from our lungs because for the change of energy $\delta E$, change of volume $\delta V$ and extra pressure $\Delta P$ (that is the difference between ...


5

Don't even think about this!! REALLY. The risk is probably low, but the outcome of a broken rope is pretty severe in climbing. Too severe for the sake of a tug of war game. This is an advanced materials engineering question. 9kN is presumably the rope's rating, right? 50-60 grownups will easily impart this force onto the rope (30 at each end, each pulling ...


5

Collisions can be elastic or inelastic. Elastic collisions are collisions where the incoming and outgoing kinetic energies are the same and only the angles change The same holds true classically, example: billiard balls ; and in the elementary particle framework. example: An electron hitting an electron has a probability of scattering elastically. ...


5

The critical parameter for materials under stress is the strain, defined in a general way{*} as the fraction change some length over that same length: $$ \lambda = \frac{\Delta l}{l} $$ So take a given a fiber of diameter $d$, and bent around a radius of curvature $r$, the strain of either the inside or outside edge is: $$ \lambda = \frac{((r \pm d) - r) ...


5

The initial and final momentum are not the same because the ball is not an isolated system. The wall exerts a force on it. In principle the ball and the wall (and the planet it's connected to!) form an isolated system with a conserved momentum, but you'd have to take into account how much the wall moves after the collision. The change of momentum is final ...


5

Hooke's Law is frequently used to model multi-dimensional materials because the stress tensor is simple (linear). The full expression can be found on Wikipedia. The simplification for 2D is straight forward (drop any terms with a 3 in the subscript). Note that whether deformation in one dimension affects the others is a property of the material and shows up ...


4

Having spent a great deal of time working with "normal" glass (one of my many hobbies), I can assure you that, in fact, all glasses can be bent. When cutting large sheets of glass, I always see the sheet bend before it breaks. Every now and then you have to strike it twice; the first time you bend it, it fails to break. Fiber optics differ from the usual ...


4

Young's Modulus isn't dimensionless! It says STRAIN is dimensionless (which is true). SO Y = Stress/Strain = [Pressure]/[Dimensionless] = [Pressure]! Young's Modulus has the same dimensions as that of pressure, which is: $[M] [L]^{-1} [T]^{-2}$ And units of pressure, which is Pascal.


4

If you use a plate glass window instead of a wall you'll find that the rubber and iron balls bounce by a similar amount (though be careful throwing iron balls at windows :-). It's a basic principle in physics that energy cannot be lost. The rubber ball starts off with kinetic energy, hits the wall, and rebounds moving with about the same kinetic energy. So ...


4

Zero strain does not always imply zero stress and visa versa. There are matterials that display stress-strain, $\sigma-\epsilon,$ hysteresis behaviour. In matterials like this, when you start loading them, they behave normally, i.e increasing the stress increases the strain. However, when you start to unload them (remove the load), instead of the stress ...


4

When a metal spring is stretched beyond it's elastic limit, the metal begins to undergo some plastic deformation. This is a permanent deformation of metal crystals caused by the creation and motion of crystal lattice dislocations. These processes are partially irreversible and some of the work performed to deform the spring is lost as heat.


3

Examples: stress is zero but strain is present= when component is loaded beyond the elastic limit it shows plastic deformation which can not be regained. after unloading the specimen in plastic deformation zone material will follow slope similar to the elastic slope and will come back to zero stress (as load is removed now). but during this process it has ...


3

Just check your math. Correct derivation is $$\begin{eqnarray} {f \over \epsilon_{11}} & = & {S_{11} \over \epsilon_{11}}=(2\mu+\lambda)+\lambda {\epsilon_{22}+\epsilon_{33} \over \epsilon_{11}} = \\ & = & (2\mu+\lambda)-{\lambda^2 \over \lambda + \mu} = {(2\mu + \lambda)(\lambda + \mu) - \lambda^2 \over \lambda + ...


3

What you need is the Euler-Bernoulli beam theory. The last three pages of this PDF explain the eigenmodes.


3

The key parameter here seems to be how are distributed the losses at each bounce (from what you said, there is no air friction here): how much translational and vertical speed are lost in the transition before/after the bounce. From your graphic, it seems like the hypothesis is that there is "reflection" of the ball: same angle when arriving/leaving the ...


3

I'd swap the meaning of x and y to make the sparsity structure more conspicuous (block diagonal). The I'd try to identify the paramaters as in the isotropic case, where they exist, and itnroduce new ones for the others by analogy (note that each column has the same denominator). Then consider special stress vectors that affect only few strain components, and ...


3

As this looks a lot like a homework question I'll give some brief hints. Consider which way the ball must be spinning. Will the magnitude of the horizontal velocity $v$ change? Whats the relationship between rotational velocity and linear velocity for a point on the circumference?


3

A toy model of a solid A very simple model of a solid is to imagine a bunch of molecules linked to their nearby neighbors by springs (you don't need to imagine a crystalline lattice, it can be amorphous). The springs are the effective electromagnetic interactions between the molecules; they are strongly repulsive at close range and attractive at modest ...


3

I believe potentials in Maxwell's equations were introduced originally to make solving equations simpler -- it wasn't until a bit later that the Lorentz symmetry was noticed. Similar potentials are introduced in 2D hydrodynamics as well, see e.g. stream function. So I would say that the reason for introducing these potentials originally was always to ...


3

If the string is massless and taut, then the wave velocity is infinite - that is, a component of the force at one mass will immediately be felt at the other mass. But to answer your first question, not all the force of the impact will be transmitted along the string, as made clear by this diagram: As for your second question: momentum for the system is ...


3

You can analize the problem by imaginarily breaking the rod and calculating the internal tension, as in the pictures below. In the first case, you get that the half between the force at the center and the force-free end has no internal stress see(1), and thus it does not elongate. The other half however, has an internal tension T' (see (3) ) of 5N at any ...



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