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2

The kinetic term of the Lagrangian is proportional to $$g_{ij}v^iv^j$$ where the $v$s are the generalised velocities. Writing them as the time derivative of the generalised coordinates, i.e. $v^i\dot q^i$, taking the square root, and multiplying by a small time lapse $\epsilon$ you get $$\sqrt{g_{ij}\dot q^i\dot q^j}\epsilon,$$ which is a first order ...


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There is this famous reality: Addition or subtraction of any constant to potential energy doesn't change the equations of motion. In your case $U_0$ is just a constant that one can add or subtract freely. You can assume that system had an initial constant potential energy (independent of your generalized coordinates of course) just before you started to ...


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As another answer points out, a constant can be added to the potential energy without affecting the equations of motion. Often, we impose the boundary condition that the potential energy is zero 'at infinity'. For the case of a central gravitational (attractive) force, imposing the "zero at infinity" boundary condition means that the gravitational ...


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You have this negative quantity because you need to pick a zero point for energy. It's a sort of needing of an arbitrary costant. But another important thing is that the system you're considering is a legacy system. Now I tell you what it is: A legacy system is a particular system where a force operate with a big power, so for separating the two objects of ...


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Well, I guess see floating things are yet to reach the ground level, because ofcourse an object submerged at the bottom most part of the seabed would have less potential energy than the object which is floating right? Implying that floating objects do have potential energy


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Negative energies are totally fine, because you had to pick a zero-point for energy. In your calculation you picked it to be at infinity. You could have chosen the zero-point for potential energy in such a way that your system had zero energy, or whatever. Only changes in energy are meaningful, in general. Consider this: what happens if you add energy to ...


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What happened with $V\left(\sqrt{x^2+y^2+z^2}\right)$? You mean, why does V(r) disappear from the $\frac{\partial }{\partial \dot q_j}$ term, right? It's because V(r) is a function only of $q_j$ not $\dot q_j$. Those variables are treated as independent and so $\frac{\partial V}{\partial \dot q_j}=0$. and why $\partial\dot q_{j} = \partial\dot ...


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How is it possible that I "invest" a (constant) teacup worth of potential energy, yet "gain" the potential energy of rising an arbitrary big mass that is floating in my basin by the constant amount caused by the water from the tea cup? The mass cannot be "arbitrarily big". Since it is floating, it has a net density that is less than that of the water. ...


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Think about a body floating in a cylinder of water. The body has a diameter that's only a little less than the cylinder. If it's a small body in a small cylinder, like your rubber duck, your teacup full of water lifts it up a lot. If it's a big body in a big cylinder, like your ship, your teacup full of water lifts it up a little. There's nothing dubious ...


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I think what he's saying is that $$F_{net} = F_{nc} + \nabla U,$$ which is pretty standard. $f^a$ is your net force, which is the sum of your conservative and nonconservative forces. Conservative forces can be written as the gradient of some potential, which is where you get your $\nabla U$ from. $f^e,$ then, are your nonconservative forces.


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As pointed out in one of the comments, you have already declared the mass $M$ (in the first case) to be fixed. Asking to prove that it indeed is fixed rather than moving is like asking to prove a definition: No need, it is already a given by power of your mind as author. I suspect you are seeming some sort of paradox. There is none: The two cases you ...


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The first case is more of a thought experiment, since gravity would move the mass M. To apply it no real life, something must be attached to M. This attached thing would move to so it must be attached to something else. This goes on ad infinitum. This is why a ball gravitationally attracts a another ball on the earth, but also the earth. Situation 1 is ...


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Energy is conserved, so you always have to ask whose energy, and where did it come from. Because the same number could appear with opposite signs in different places. So we know the kinetic energy, it's the kinetic energy of the unnamed particle experiencing all these named forces. But what about the potential energy, where did it come from? Let's name ...


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What if I heat up the system? Where does this extra energy ($W_{ext}$) go? EDIT: could $\Delta U$ be the change in potential? By introducing some net amount of work, you could move two electric point charges closer together, for instance.


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Based on the answers here is my summary: Kinetic energy of one plane: $$ E_{kin} = 2 * 10^9 J = 0.5 \text{ kilotons of TNT} $$ Chemical energy of the plane fuel (very close to the amazing estimation of Floris): $$ E_{chem} = 38000 L * 35 * 10^6 J/L = 1.33*10^{12} J = 330 \text{ kilotons of TNT} $$ Potential energy of one collapsing tower: $$ E_{pot} = ...


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The $\frac{1}{2}$ is missing in their derivation because the potential you are both calculating is different. You integrate from $0$ to $y$. In here is the implicit assumption that at the equilibrium height there is zero potential. In the derivation you linked, they use the formula $U = \Delta mgy$. What they are doing is taking a column of fluid of height ...


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I figured it out. It's a really horrid approximation that's only valid for small $y'$ and doesn't seem grounded in mathematical rigor. It's only reasonable because as soon as you get any appreciable buckling ($y' \neq 0$), the rod collapses anyway. The arc length of the rod is: $ \int_0^L ds = \int_0^L \sqrt{1+y'^2} dz \approx \int_0^L 1 + \frac{1}{2} y'^2 ...


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Assuming kerosene is C8H18, has 25 chemical bonds, each of which releases 1eV when burned, gives an energy in fuel of 20 MJ/kg. The weight of a plane shortly after take off is significantly fuel. If I'd guessed I would have said 20 tonnes per plane; (Floris' comment suggests more like 30). This gives 400 GJ per plane, equivalent to 100 tons of TNT per ...


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Estimating is always a fun aspect of physics - so let's do some, without looking up any values. What is the kinetic energy of a plane? We need to know the mass of a plane and its speed. I am going to use seriously rounded numbers - let's see how close we get. We "know" a full size car is about 1000 kg, and can carry 5 passengers of 100 kg. That means a car ...


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The Poynting formula for electrostatic energy in volume $V$ $$ E = \int_V \frac{1}{2}\epsilon_0 E^2 dV $$ can be derived from the Coulomb law only for cases where the field acting on the particles is defined everywhere. However, point particle has infinite charge density at the point it is present and the field is not defined at that point. So the ...


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Aircraft fuel is kerosense. Sometimes it is difficult to directly measure the amount of heat something produces. We can make the process easier by burning an amount of the fuel to heat water. The energy lost by the fuel can then be calculated by finding the heat gained by the water as measured by the change in temperature of the water. this reference ...


1

It depends on how you define the potential. In mechanics one has the convention $$ F = -\nabla U$$ Since the electric field exerts a force via $F=qE$ it is only natural to apply this convention in electrostatics too. In this way the electric potential $V$ can be directly interpreted as mechanical potential energy $U=qV$. Option number (2) is therefor the ...


0

The following is basically what the Ashcroft/Mermin says about it. The idea is as following: in harmonic approximation a relative displacement $u$ results in an energy $U=- \frac 1 4 (\vec{ u }(\vec R) - \vec{ u }(\vec R ')) \mathbf{D}(\vec{ u }(\vec R ') - \vec{ u }(\vec R)) $ The tensor $\mathbf D$ already has natural symmetry ...


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You should start with the strain energy density $\psi$, then define: $$ C_{ijkl} = \frac{\partial^2 \psi}{\partial \epsilon_{ij}\partial \epsilon_{kl}}, $$ and then define $$ \sigma_{ij} = C_{ijkl} \epsilon_{kl} $$ The remainder of my answer will be about explaining why you have to do it that way. Firstly it is physical, there really is energy associated ...


1

Since $\epsilon$ is a symmetric tensor, it has 6 independent component that determine it. Hence use a multi-index $I\in\{(i,j)|1\leq i\leq j\leq 3\}$ to denote them. The strain energy density then becomes (perhaps one has to be careful with "diagonal" terms here in order to get the right coefficients) $$\psi = C_{IJ}\epsilon_I\epsilon_J$$ where summation is ...


0

The notion of negative energies is more like a convention, since potential energy is defined up to an arbitrary constant. In the case of the Coulomb potential it is assumed as a convention that the potential generated by a charge is zero at infinity, where there would be no interaction with other charges. Energy levels which are then lower than this one, ...


1

The potential energy of the ball is given as $E = mah_1$ where $m$ is the mass of the ball, $h_1$ is the height over the point you set as zero potential and $a$ is the acceleration due to gravity (which is different between Mars and Earth). If you assume there is no energy lost due to heating of the ball or other inelasticities, you have an elastic collision ...


3

In general, when working out the Lagrangian, start in coordinates that you know and then rewrite in generalized coordinates. Kinetic energy in this case is proportional to $v^2 = \dot x^2 + \dot y^2 + \dot v_z^2$. In your spherical coordinates $$x = r \sin \alpha \cos \theta;~ y = r \sin \alpha \sin \theta;~ z = r \cos \alpha.$$ Take full time derivatives ...


1

The kinetic energy of the particle depends just on the path it is following - if you imagine the cone is suddenly invisible, the particle continues to go around in a circle. That means that there is no reason to add $\sin\theta$ in your expression for the kinetic energy if you used $r$ to mean (as drawn) the distance from the axis of rotation. Note - your ...


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There's no such thing as an imaginary point. In other words, you can certainly plug imaginary numbers into a formula, but those imaginary numbers don't represent points in space, and thus the results you get will not represent the conditions at any actual point. However, you don't need an actual point in space for this purpose. The potential energy field ...


0

You can't state what the potential energy of a system is in absolute terms, you can only talk about the difference in potential energy between two states. In your example, you are quite free to define the potential energy of two nearby particles having the same charge as anything you like, positive, negative or zero. For example, what is the potential ...


2

It is up to you where to define the zero potential (potential energy undefined up to a constant, so by adding any constant, the zero becomes anywhere you want it), but let's consider the point where the potential itself is minimum and subtract this value so that the potential is everywhere positive except at it's minimum (just a convention) $$\vec{F} = ...


0

There was a mistake in the Integration: $- \int_{x_i}^{x_f} F dx = [\frac{A}{x}+\frac{1}{2}Bx^2]_{x_i}^{x_f}$. When you set the potential difference $U(x_f)-U(x_i)$ to Zero, you will get an $x$ that is always a real number; you will have $x = (\frac{-2A}{B})^{\frac{1}{3}}$. Usually, imaginary potentials are unphysical; every measurement of physical ...



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