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11

The "simplest" classical explanation I know is the van der Waals interaction described by Keesom between two permanent dipoles. Let us consider two permanent dipoles $\vec{p}_1$ (located at $O_1$) and $\vec{p}_2$ located at $O_2$. Their potential energy of interaction is: \begin{equation} U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = ...


11

Of course it has something to do with the liquid water entering the gas phase just above the cup of tea, but how does that give the bag of tea a directed motion to one side? Nope. The teabag is dangled by a string. Remember that the string is made of wound up threads: Now, the threads stay wound up because they fit well and they have a knack of ...


10

Here's another way of looking at it. Let M1, M2, M3 be our three masses. In the three body problem we're considering, the whole frame containing M1, M2 and M3 is rotating. You're right to think that if that frame was fixed then the points L4 and L5 would not be stable. After all if you perturb M3 from L4 or L5 then it should just roll down the potential ...


9

The formula you quote does not contain the potential energy, it is valid for a free particle (i.e. a particle which is not affected by external potential). You can link it to classical mechanics by evaluating it for small values of $p$ (more precisely: $ p \ll c$): $$ E = \sqrt{\left(mc^2\right)^2 + p^2 c^2} = c \sqrt{m^2c^2 + p^2} = \cdots $$ $$ \cdots = ...


8

The potential energy only being defined up to a constant does not imply that potential energy differences only depend on differences in position. To see this mathematically, assume that a function $U$ has the property that $U(x_2)-U(x_1) = f(x_2-x_1)$ for some function $f$. Then if we take $x_2 = x+\Delta x$ and $x_1 = x$, and divide both sides by ...


8

Gravity is doing that work! If you observe, the domino is in a position of unstable equilibrium. Edit: as pointed out in the comments, this position is of a metastable and not unstable equilibrium. This means that the domino is in a state where it hasn't achieved the minimum possible energy state yet. The energy I'm talking about here is the ...


7

While it may be possible to derive a violation of energy conservation due to intersecting equipotentials, there is a much more intuitive and in my opinion a more fundamental reason that equipotentials cannot intersect: Potential is a single-valued function. A good analogy for potential in this case is a map of the ground elevation of the earth; a ...


7

You're right that if you take Newton's law of gravity as is and apply it to a 2D universe, you'll get an infinite result. So you do need to use a modified theory in two dimensions, or indeed in any number of dimensions other than three. The proper way to do this is using general relativity, and if you apply GR to 2+1D spacetime, you get something that looks ...


7

The energy in your equation is for a free rigid body in the absence of a potential. We can see this if we start with a Lagrangian with a scalar function, $\Phi(q)$, and remember $\gamma$ is a function of $\dot{q}$, $$ L=T-V=-\gamma^{-1} (\dot{q}) \, mc^2-\Phi(q) $$ Then if we find the momentum $$ \pi=\frac{\partial L}{\partial ...


7

Let $E$ denote a quantity that does not change over time (from the first principle). Consider a ball with mass $m$ dropped from a height $h$. As the ball drops, its speed changes due to the gravitational acceleration $g$, reaching a final value $v$ at impact. Thus, we can infer that the quantity $E$ depends on these 4 parameters: $$E(m,H,g,V)$$ where $H$ ...


6

Yes the free body moves outward, but there are two critical things you have to know to interpret this statement correctly. First, this is the effective potential, taking into account gravity and centrifugal force. It has this form because we went into the non-inertial frame co-rotating with the two masses. Mathematically, the potential is $$ ...


5

Yes, u is indeed the potential energy. And yes, you can calculate the force acting on a particle by calculating the gradient of the potential energy field at the position the particle is in. Computationally you will want to calculate the force on particle 1, by taking the gradient at the position particle 1 is in, of the potential energy field created by ...


5

Special relativity doesn't alter the fact that interactions between particles "store energy" in the form of "potential energy," alhtough special relativity does alter the terms you listed, all of which have to do with the energies possessed by particles either by virtue of their motion, or their mass. For example, in special relativity, electromagnetic ...


5

It's valid in the sense that it does tell you the rest energy of a 200-pound person, but it does not tell you how much energy you could get by splitting all those atoms. As a matter of fact, most of the atoms in a human body are carbon, nitrogen, and oxygen; splitting these atoms takes energy, it doesn't produce it. Your character would need to tap into a ...


5

Think about the work-kinetic energy theorem, which states that the net work done on an object is equal to its change in kinetic energy: $$W_{net}=\Delta\mathrm{KE}.$$ You are right that when lifting an object of mass $m$ by a height $h,$ in a uniform gravitational field, the work you do is $W_{you}=mgh$ (assuming, as you said, that you're applying a force ...


5

Your teacher's explanation is incorrect. A simple counterexample can be constructed to illustrate this by considering what happens when the role of your arm is replaced by that of a rubber band. When a weight is suspended from the ceiling by a rubber band, the band stretches and its polymer chains become more ordered, in exact analogy to your teachers ...


4

You've got basically the right idea. Just for clarity, let me recap the setup: suppose that your ring is centered at the origin and oriented in the xy plane. Consider two differential elements of charge, $\mathrm{d}q$ located at $(R,0)$, and $\mathrm{d}q'$, located at $(R\cos\phi,R\sin\phi)$. The potential energy of these two charge elements is ...


4

It is indeed correct that only the difference between two potential energies is physically meaningful. An in-depth explanation follows. For the rest of this answer, forget everything you know about potential energy. I suppose you know that when you have a conservative force $\vec{F}$ acting on an object to move it from an initial point $\vec{x}_i$ to a ...


4

Yup. Inside the (uniform spherical) mass, IIRC $\phi=-\frac{GM}{2R^3}\left(3R^2-r^2\right)$. Or something like that. So, $$\phi=\begin{cases} -\frac{GM}{r}, & r>R \\ -\frac{GM}{2R^3}\left(3R^2-r^2\right), & r<R \end{cases}$$ The laplacian $\nabla^2\phi$ should be $$4\pi G\rho=\nabla^2\phi=\begin{cases} 0, &r>R \\ 4\pi G\rho_0 ...


4

As I know, in L4,L5 , the potential of the Gravitational power is at it's maximum, although it is unusual for kinematics, that considers stable points to be when U->Min , but in Dynamic systems, stable points can be even when U->Max , but then we call it "Dynamical Equilibrium" in sense that the object will actually move around the stable point (but will ...


4

When you look at the dynamics in the rotating reference frame, there are 4 forces acting on the particle: the two gravitational pulls from the massive bodies, the centrifugal push away from the center of rotation (located between the massive objects) and the Coriolis force. The first three forces depend on the position of the particle, and can be derived ...


4

There is no need for any empirical evidence. This is pure mathematics. Step 1: Assume a force is conservative. This means that ${\vec \nabla} \times {\vec F} =0$ Step 2: Then, via Green's theorem, you know that the quantity $\int_{a}^{b}{\vec F}\cdot d{\vec s}$ does not depend on the path you take from a to b. (equivalently, this integral is zero if ...


4

The comparison is viable, here's why: Let's choose the positive $x$-direction to point upward, perpendicular to the water's surface. By Archimedes' principle, the magnitude of the buoyant force on an object of volume $V$ equals the weight of the displaced water; $F_B = \rho_w V g$ where $\rho_w$ here denotes the density of water. The buoyant force ...


4

Yes, the derivative of a step function is a Dirac delta. You can see this by integrating the delta function: $$ \Theta(x)=\int_{-\infty}^x \delta(x') \mathrm{d}x'$$ where $$\Theta\left(x\right)=\begin{cases} 1 & x>0\\ 0 & x<0 \end{cases}$$ (note that $\Theta(0)$ is not defined by this prescription. If you use a symmetric representation of the ...


4

You can actually do this using $\delta$ functions as well if you'd like (provided you're careful). Let $V(x) = V_0\theta(x-x_0)$, then $$ F(x) = -V'(x) = -V_0\delta(x-x_0) $$ For simplicity, let's take the potential barrier to be located at $x_0 = 0$ so that by Newton's second law, the equation of motion becomes $$ -V_0\delta(x(t)) = m \ddot x(t) $$ ...


4

No, most force fields refuse to be conservative. The path-independence is a nontrivial constraint in an arbitrarily small region of space, an arbitrary neighborhood. If the force field is conservative, it must be $$\nabla\times \vec F = 0$$ because $F = -\nabla\Phi$. It's clear that the curl of $\vec F$ may be nonzero even if you look at a small ...


4

Yes, it is a torsion spring. It works by twisting the metal rod that makes up the body of the spring. The reason for coiling the spring is to fit a long length of metal rod into a short space. You need a long length of rod so that the torsion per unit length remains small. With a shorter length of rod you'd exceed the elastic limit and the rod would be ...



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