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72

Your wire is not quite round (almost no wire is), and consequently it has a different vibration frequency along its principal axes1. You are exciting a mixture of the two modes of oscillation by displacing the wire along an axis that is not aligned with either of the principal axes. The subsequent motion, when analyzed along the axis of initial excitation, ...


64

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 ...


32

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 ...


21

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 from the uninflated state (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 ...


16

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 ...


16

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 ...


13

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.


12

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 ...


12

UPDATE : After looking again at the video, I agree that Floris' explanation seems to be correct and my explanation below is wrong. Slightly different frequencies of vibration in two perpendicular planes accounts more simply for a rotation which reverses one way then the other. Kinetic energy seems to decay constantly; it does not seem to be stored in an ...


7

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 ...


7

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. ...


7

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. ...


7

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 ...


6

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) \...


6

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 ...


6

Here's some tennis racket physics from Rod Cross, including links to several Am. J. Phys articles (the physics educators' journal, thus excellent for learning from) and this excellent diagram: There are at least three "sweet spots": The node, at the center of the strings, is a point where the natural standing waves in a vibrating racket don't have any ...


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

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

You can think of two atoms in a solid joined together via a bond. The graph of potential (pe) energy against separation of the atoms is shown below as is a graph of the force between atoms and their separation. The equilibrium separation $r_o$ is when the potential energy is a minimum and the force between the atoms is zero. Now it so happens that foe ...


5

Will a tennis ball go further if i hit it with the side of the racket? No. You want the racket to deform, not the ball. This means using the strings to elastically store energy and return it to the ball. The Ball The ball's deformation upon impact is undesirable because "a tennis ball is required by the rules of tennis to dissipate a fraction of ...


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

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

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 ...


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.


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

To state the physics of this question: fire raises the temperature of load bearing steel. The latter's mechanical properties radically change - yield strength in particular drops radically - leading to structural failure under normally well bearable loads. The engineering stuff: Steel is a known, severe fire hazard in this way, but its strength and low ...


4

The basic idea is, that a condensed matter system can be described by a collection of particles, that have pairwise interaction forces. These can be written as derivatives of a potential $V(\vec r_1, \ldots, \vec r_n)$. (Technically, we get such a potential for the nuclei, which move much slower than the electrons due to their higher mass, by doing a Born-...


4

The elongation, or shortening of the materials with temperature depends on the coefficient of linear thermal expansion of the material with which it is made (here: gold) $$\alpha = \frac{1}{L}\frac{dL}{dT}$$ Integrating, $$L = L_0\ e^{\alpha\Delta T}$$ $$L \approx L_0(1+\alpha\Delta T)$$ In your case, $L$ = circumference. Change in radius will be given by: ...


4

Multiply both sides of the equation by $\Delta l$ to obtain: $$k\Delta l=EA\frac{\Delta l}{l_0}$$ The relationship on the left is the tensile force F. The quantity $\frac{\Delta l}{l_0}$ is the tensile strain. The Young's modulus E times the tensile strain is equal to the tensile stress $\sigma$:$$\sigma=E\frac{\Delta l}{l_0}$$ The tensile stress times ...



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