# Tag Info

20

Imagine a rock on a rope. As you rotate the rope faster and faster, you need to pull stronger and stronger to provide centripetal force that keeps the stone on the orbit. The increasing tension in the rope would eventually break the it. The very same thing would happen with bar (just replace the rock with the bar's center of mass). And naturally, all of this ...

16

I remember that the question in your title was busted in Mythbusters episode 72. A simple google search also gives many other examples. As for single- vs alternate-direction folding, I'm guessing that the latter would allow for more folds. It is the thickness vs length along a fold that basically tells you if a fold is possible, since there is always going ...

15

Consider a standard volume of $1\textrm{ m}^3$ of air. This contains on the order of $10^{25}$ molecules of O2 and N2. If you needed to simulate or explain the physics occurring in that volume of air, would you want to model $10^{25}$ molecules and all the interactions between them or, say, 100x100x100 cells based on the Navier-Stokes equations? ...

14

There is a real object with relativistic speed of surface - millisecond pulsar. The swiftest spinning pulsar currently known, spinning 716 times a second. Surface speed of such pulsar with radius 16 km is about $7*10^7$ m/s or 24% speed of light. It is calculated that pulsars would break apart if they spun at a rate of more than 1500 rotations per ...

8

You are observing a hydraulic jump. The Wikipedia article is very good, so I won't try to out-do it. In brief summary, when the water starts running out from the place where it hits the sink, the same flux is spread out over a larger and larger circumference as you move out. This means the flow gets shallower and moves more slowly as you move further ...

7

Are you referring to the exact relativistic equivalent to Navier-Stokes equation or a more general Dissipative Relativistic Hydrodynamics Equation? The "relativistic equivalent to Navier-Stokes equation" would be something like this: There would be an energy momentum tensor with the following form: $T_{\mu\nu} = (e+p)u_\mu u_\nu - p g_{\mu\nu} + ... 6 What does it mean? The reason they are conservative or non-conservative has to do with the splitting of the derivatives. Consider the conservative derivative: $$\frac{\partial \rho u}{\partial x}$$ When we discretize this, using a simple numerical derivative just to highlight the point, we get: $$\frac{\partial \rho u}{\partial x} \approx \frac{(\rho ... 5 Hooke's Law is frequently used to model multi-dimensional materials because the stress tensor is simple (linear). The full expression can be found on Wikipedia. The simplification for 2D is straight forward (drop any terms with a 3 in the subscript). Note that whether deformation in one dimension affects the others is a property of the material and shows up ... 5 The onset of turbulence in fluids is determined by the Reynolds number$$ \mathrm{Re} = \frac{vL}{\nu}, $$where L is the characteristic length scale, v the characteristic velocity, and \nu the viscosity. The onset of turbulence in fluids occurs for \mathrm{Re} greater than about 1000 or more, depending on geometry. If we want to see the equivalent ... 5 Indeed, both the strain tensor$$\epsilon_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right) \tag{1}$$and the stress tensor$$\sigma_{ij}=2\mu\epsilon_{ij}+\lambda\epsilon_{kk}\delta_{ij} \tag{2}$$are symmetric by definition. However, bear in mind that these definitions are not always valid; (1) assumes ... 5 The momentum flux tensor comes from the momentum equation of Navier-Stokes equations:$$ \frac{\partial\left(\rho\mathbf{u}\right)}{\partial t}+\nabla\cdot\mathbf{P}=0\tag{1} $$Or, using indices (where it is easier to see that \mathbf{P} is a rank-2 tensor):$$ \frac{\partial\left(\rho u_i\right)}{\partial t}+\frac{\partial\Pi_{ij}}{\partial x_j}=0\tag{2} ... 4 remember that in a three-dimensional description of special relativity the impulse of an object is given by $$\mathbf{p} = \gamma m \mathbf{v}$$ with the so-called Lorentz-factor $$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$ Now, do you think you can accelerate the masses within the slab to a speed greater than light or do you think that something is wrong with ... 4 I walked into my local Mech. E. department and had a chat with a continuum mechanist. Two books he showed me that looked like pretty good answers were: Marcelo Epstein and Marek Elzanowski. Material inhomogeneities and their evolution: a geometric approach. Marcelo Epstein. The geometrical language of continuum mechanics. One thing I like about ... 4 There are many physical intuitions often presented in various texts on fluid dynamics. I won't mention those here. I will, however, mention that mathematically the passage from a particle point of view to a continuum point of view is still a largely un-resolved problem. (With suitable interpretation, this problem was already posed by Hilbert as his 6th of 23 ... 4 The fabric covering the foam pad has a warp and weave (we can assume.) The fabric can stretch with the warp or the weave but not at a 45 degree angle, called the bias direction in sewing. So, when the ball hits the fabric it causes a wave in the fabric which begins to travel outward like a ripple in water. The wave causes distortion of the fabric as it ... 4 It provides a convenient graphical means of finding the maximum and minimum shear stress, which are important for determining material failure. You don't absolutely need it, but the graphical interpretation of the circular relationship between normal and shear stress is somewhat convenient. I've read good solid mechanics books that give little if any ... 3 The theory of fluids introduces material parameters in the stress tensor, which help model the substance. "The viscosity coefficient is the proportionality constant relating a velocity gradient in a fluid to the force required to maintain that gradient. The thermal conductivity is the proportionality constant relating the temperature gradient across a fluid ... 3 I say no. Assuming all the practicalities work, you can get arbitrarily close to c. But not reach c. You can see this easily from the relativstic formula for kinetic energy:$E_k = mc^2(\frac{1}{\sqrt{1-v^2/c^2}}-1)$As$v$approaches$c$, the energy you need to supply to a particle at the end of the bar tends to infinity. 3 Approximatation:$a(n)$- Area after$n$folds.$t(n)$- Thickness after$n$folds.$d(n) = k\frac{t(n)}{d(n)}$- Difficulty after$n$folds.$a(n+1) = \frac{1}{2} a(n) \rightarrow a(n) = c_1 2^{1-n}t(n+1) = 2 t(n) \rightarrow t(n) = c_2 2^{n-1}d(n) = k4^{n-1}$Physics: You can't fold an atom. Area of atom$= a(N) = c_1 2^{1-N}$Solve for ... 3 An interesting question. You are right, the stress in a crystalline solid, or any solid, is treated by engineers as a macroscopic property of matter assuming matter is a continuous medium. It is given in terms of the external forces acting on the solid per unit area at some direction. Hence the distinction of$\sigma_{xy}, \sigma_{yz}$etc. This goes with ... 3 What you need is the Euler-Bernoulli beam theory. The last three pages of this PDF explain the eigenmodes. 3 Landau and Lifshitz, Theory of Elasticity, Course of Theoretical Physics, Vol 7 (1986) An older edition is available online here. In my 3rd edition printing, the very same equation (well, with slightly different notation) is on page 92. In the online edition I referenced, check page 107. 3 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 Speaking about Cauchy stress tensor in classical mechanics, the answer to your first question is that it does not matter, as you have metric in arbitrary coordinates induced by dot product of underlying Euclidian space. You can exploit symmetry of Cauchy stress tensor from balance of angular momentum assuming no couple-stresses, i.e. sources of angular ... 3 Just calculate the difference between the first and second equation you want to show to be equivalent. The difference should be$0=0$if they're equivalent. You actually get: $$-\rho \frac{D}{Dt}(\frac{v^2}{2}) =\sigma:D-\nabla\cdot (\sigma\cdot v)-\rho f\cdot v$$ All the other terms agree. The equation above really says$0=0$because the terms may be ... 3 The Cauchy stress matrix$\Sigma$is a$3 \times 3$real symmetric matrix. It is interesting that we may without problems generalize$\Sigma$to a$3 \times 3$Hermitian matrix. It has three mutually orthogonal principal stress directions with principal stresses (eigenvalues)$\lambda_1\geq\lambda_2\geq \lambda_3$. Consider an arbitrary unit vector ... 3 Functional derivative:$F_d[{\bf p}] = \int \mathrm d\boldsymbol{r}\ f( \boldsymbol{r}, {\bf p}(\boldsymbol{r}), \nabla\cdot {\bf p}(\boldsymbol{r}) )\frac{\delta F_d}{\delta {\bf p}(\boldsymbol{r})} := \frac{\partial f}{\partial {\bf p}} - \nabla \cdot \frac{\partial f}{\partial\nabla\cdot {\bf p}},$although something's weird with your signs. Your ... 3 Although I am not 100% sure what you are asking, I believe you are talking about considering a lattice of discrete points and taking the continuous limit such that the distance between the lattice points tend to zero. The simplest example is to consider an 1-dimensional elastic rod with mass density$\mu$. The force applied on it due to Young's modulus is ... 3 The formula that you write out is just a consequence of linearity without any additional requirements. Suppose you had a vector$\mathbf{D}$that was linearly dependent on another vector$\mathbf{E}$. Then, one would write$D_i =\sum_j \epsilon_{ij}\ E_j$. With no further conditions,$\epsilon_{ij}$would have$9=3\times3\$ coefficients. In your formula, the ...

2

Dear Hernan, as the distant parts of the bar are approaching the speed of light, they become heavier, so it becomes harder to accelerate them: you can never reach (or surpass) the speed of light. It doesn't matter whether you try to accelerate the "final segments" of the bar by jets or by their attachment to the rest of the bar that is being pushed in the ...

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