# Tag Info

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The explanation comes from earlier in that paragraph: If all the co-ordinates and velocities are simultaneously specified, it is known from experience that the state of the system is completely determined and that its subsequent motion can, in principle, be calculated. This is just saying the familiar thing that if you know the laws of physics for the ...

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The rate at which the wheel is spinning is an angular velocity, normally measured in radians/second. The velocity of the cart is a linear velocity - metres/second in SI units. Since the two are different units, they can't really be compared. However, you can calculate the instantaneous velocity of any given part of the wheel rim. Unless the wheels are ...

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If I consider the comment of fall-apart: To explain the behavior: as an epiphenomenon caused by a stimulus everywhere on the BM in combination with the resonance frequency distribution along the BM. High near the RW and low near the helicotrema. And that epiphenomenon is nothing else but a phase wave. The crowd on the gallery in a stadium ...

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The definitions are equal: Sum of external forces zero, sum of external torques is zero. This comes from classical mechanics. For a perfect ideal fluid, the external force density is the pressure gradient: $\mathbf f = -\nabla p$, and therefore, uniform pressure in a fluid means no external force on it, and then it is in mechanical equilibrium. So, its more ...

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Yes. The manner of which two surfaces in contact interact is highly investigated by the Tribology community.In particular, the field exploring the mechanics of the interaction is called contact mechanics. Tackling problems of contact mechanics analytically/numerically is often done by solving the elasticity equations. By predicting quantitatively the forces ...

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The OP has a point. If a dot denotes time-differentiation $$\dot{q}~\equiv~ \frac{dq}{dt},$$ and if we add a total time derivative to the Lagrangian $$\tilde{L}(q,\dot{q},t)-L(q, \dot{q}, t) ~=~ \frac{dF(q,t)}{dt}~\equiv~\frac{\partial F(q,t)}{\partial q}\dot{q} + \frac{\partial F(q,t)}{\partial t},$$ and if we want to view position $q$ and velocity ...

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Rigorously speaking, yes, you are right if dealing with the Lagrangian function. Indeed E.-L. equations should be more properly written $$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}^k}\right) - \frac{\partial L}{\partial {q}^k}= 0\:, \quad \frac{d q^k}{dt} = \dot{q}^k\quad k=1,\ldots, n\:.$$ In other words $\dot{q}^k$ becomes $\frac{d q^k}{dt}$ ...

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p dot in d'Alembert's formula represents the derivative of the object's momentum with respect to time. delta * r represents what would have been the displacement of the object during the infinitesimal interval of p dot. So what d'Alembert is saying is that Force minus the effect of Force = zero. If you analyze a system this way, you can treat it as though ...

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The momentum is a covector because it is a gradient, and gradients are always covariant. It does what it says on the tin. However, you are right that this is a subtle point and it's not particularly clear at first sight. For a lagrangian of the form $L=T-V$ with $V$ independent of $\dot q$, the canonical momentum is given by $$p=\frac{\partial L}{\partial ... 3 Dimensional analysis only gives results that are correct to first order and up to a constant. So really there is just some qualitative guessing on physical behaviour with dimensional analysis. 2 Yes it can! However, the term dimension analysis needs to be seen in its/a context. Buckinghams Pi-Theorem did not just emerge out of the blue. And, the reason it works is not pure luck. There is a well grounded physical/theoretical basis for it. And it is the basis which allowed for dimesion analysis. Three scientists come to my mind. Sophus Lie wrote ... 4 As an addendum to @CuriousOne answer, also in mathematical analysis the dimensional analysis (or more properly scale transformations) can be used to guess a priori estimates and useful results, that anyways has to be proven in a rigorous fashion by other means. Two relevant examples may be the Sobolev and Strichartz estimates, whose admissible indices can ... 12 Dimensional analysis can help to "guesstimate" the form of many important results but it can, for instance, not produce general solutions to equations of motion. It's an invaluable tool to understand the structure of physical theory, including quantum mechanics and relativity, and to check results for consistency, but it can rarely replace complex ... 4 Fundamentally, this is no different from computing the friction in a fluid (shear viscosity). The theory of viscosity goes back to Maxwell and Boltzmann, and microscopic calculations are possible for many fluids. Solid friction is more complicated, because the exact preparation of the surface obviously matters. First principles theories therefore concentrate ... 0 See http://ruina.tam.cornell.edu/Courses/ME4730%20Fall%202013/Rand4770Vibrations/BeamFormulas.pdf$$ \delta = \frac{5 w \ell^4}{384 E I} $$Stiffness is k=\frac{F}{\delta}$$k = \frac{w \ell}{\delta} = \frac{384 E I}{5 \ell^3}$$1 The differential amount of force is given by:$$ dF = \Delta p* dA $$You do need to add up the vertical components, which are given by:$$ dF_\downarrow(\theta) = \Delta p * dA * \sin(\theta) $$This goes from 0 \leq \theta \leq \pi. At each value of \theta the differential area is the same:$$ dA = rd\theta * dx $$This gives that:$$ ...

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Angular momentum is that which is conserved under rotations. Equivalently, the angular momentum operators are the generators of rotations. This holds both classically and quantumly by (versions of) Noether's theorem. Defining "angular momentum" as $\vec x \times \vec p$ classically and then showing that it is conserved is doing it the wrong way around from ...

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It means there is an asymptote for $$E=U_{eff}(\theta)$$ which corresponds to a certain value of $\theta$ you can evaluate by solving the equation: $$E=\frac{M_z^2}{2ml^2}\sin^2{\theta}-mgl\cos{\theta}$$ So $E$ takes this value in a infinite time. This equation is quadratic in $\cos{\theta}$ also because it's written that there are 2 roots. Maybe it's a ...

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Since the mass is at equilibrium, the force due to rotation which depends on $m,\omega,r$ acting in the radial direction should be equal to the force due to the spring which depends on $k$ and $x$. The springs can be thought to be composed of 2 springs in parallel. Equate them to find $x$

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I won't give you the answer but to get you started, use Hooke's law and the centrifugal force expression in terms of the angular velocity ω. Remember: The centrifugal force depends on the distance of the mass from the axis, x The restoring forces provided by each of the springs are actually the same as each other since one is compressed and the other ...

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As CuriousOne inexplicably said in the comments, but not as a formal answer, you should use this equation: $$\vec{L}=m{\vec{r}}\times{\vec{v}}$$ This is the standard equation for angular momentum in vector form. Once you have your angular momentum vector, you can get the individual components. You can see how to take a cross product here. If you need to ...

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After thinking about Nick P's answer and re-reading the relevant chapter of Sussman's Structure and Interpretation of Classical Mechanics, I came up with the following elaboration of Nick's argument. It's not water-tight, but it convinced me, and perhaps it will help someone else. I will use Sussman's unorthodox but precise notation. The first step (and ...

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For simplicity consider the 1-d case, with $\psi =\sqrt{n} e^{2i\phi}$, then $$i \psi_t =\frac{i}{2} \frac{\dot{n}}{\sqrt{n}} e^{2i\phi} -\sqrt{n} e^{2i\phi} 2\dot{\phi}.$$ Similarly $$\frac{\partial H}{\partial \psi^*} = \frac{\partial H}{\partial n}\frac{\partial n}{\partial \psi^*} + \frac{\partial H}{\partial \phi}\frac{\partial \phi}{\partial ... 0 since the particle is accelerating and is charged, it should release radiation. Is this correct? This is the way experiments with radiation are often explained - radiation comes from places where acceleration of charge occurs. Also, orbital motion of charged particles in a cyclotron was found to be always accompanied with EM radiation coming off ... 0 It is possible to interpret D'Alembert's principle through the requirement that any particle is always in equilibrium in its own rest frame; it is, after all, at rest in this frame. However, as this frame is necessarily accelerating with respect to any inertial frame, there is an additional inertial force -m\ddot{\mathbf x} on the particle. The requirement ... 3 Let's take the canonical commutation relations (CCR), in their exponentiated form (Weyl's relations):$$V(\eta)T(q)=e^{-i\eta\cdot q}T(q)V(\eta)\; ,$$where \{V(\eta)\}_{\eta\in \mathbb{R}^d} and \{T(q)\}_{q\in \mathbb{R}^d} are objects of a given normed algebra with involution. This is a very general notion, that is nowadays taken as the definition of ... 0 It will decelerate causing its speed to decrease, and because of r=mv/qB the radius will decrease as well and you will get a spiral motion. This deceleration due to radiation is known as the Abraham-Lorentz force of radiation backreation. Using these equations you can more precisely derive the spiral motion. This effect is also responsible for the failure ... 1 There is, in a sense, a way to 'guide' oneself to the equations of motion based on the symmetries. The form of mechanics most suitable for this purpose is Hamilton's principle - the system takes a path for which the action has a stationary value for variations with fixed endpoints:$$\delta S=0$$S is generally expressed as (under some parametrization of ... 2 Hello @abcdef this is a very interesting problem As of today no (general) analytical solution for the NS-Equations has been found, in fact we would all hear about it, since it is worth one million dollars (see Millennium Prize). And, even though it has not been proven, there is a lot of evidence that the NS-Equations do describe fluid motion as long as the ... 1 A constraint condition can reduce the DOF of the system if it can be used to express a coordinate in terms of the others. This can always be done in case of holonomic constraints which are basically just algebraic functions of the coordinates and time. This means that you just have to manipulate the constraint equation in such a way that one of the ... 1 In particular, I want to know if the fact that accelerated charges radiate (Larmor's formula) can be derived from the Hamilton's equation of the system. If the Hamilton's equation include the electric and magnetic fields as dynamical, then yes, it should be do-able... However, if you are just including the electrostatic interaction between the ... 3 The classical limit of quantum theories is understood quite well from a mathematical standpoint nowadays. The so-called semiclassical analysis covers the QM (finite dimensional phase-space) cases, the Hepp method and infinite-dimensional semiclassical analysis cover the systems with classically infinitely many degrees of freedom. The ideas can be summed up ... 0 The so called Pulse and Glide is a driving technique that is reported to save quite much fuel. It consists of moderate accelerations (that better exploit the petrol engine) interleaved with moderate decelerations (that save fuel). This is even more effective with hybrid vehicles which can completely turn off the petrol engine without issues like the absence ... 3 I agree with user3823992 that it was incorrect to neglect the pressure differential. With the steady sinusoidal body force that's given, it's basically a hydrostatics problem with the pressure differential balancing the body force. Consider the Navier-Stokes momentum equation:$$ \frac{\partial \mathbf{v}}{\partial t}+(\mathbf{v} \cdot ...

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In principle, neglecting rolling friction and air friction, and neglecting the fact that the hilly road is longer, it should make no difference. On the other hand, if on the hilly road you ever have to either hit the brakes, or use engine braking, because of too much speed, that will cost a lot of energy. Also on the hilly road, if you have to downshift ...

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A nice answer to this question may be found in the 2009 MIT Course on the Physics of Energy (Lecture 3) by R. Jaffe and W. Taylor. They work out the energy balance of car transport as an example of mechanical energy and its conservation. Let me briefly recap their findings in my own words. The two "towns" they consider are Boston and New York, both assumed ...

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You state the second law as : The entropy of the universe always increases. In my college textbook it is stated as : Processes in which the entropy of an isolated system would decrease do not occur, or, in every process taking place in an isolated system, the entropy of the system either increases or remains constant.( F.W.Sears an introduction ...

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...two roads of the same length. One road is flat, the other road goes up and down some hills. Will an automobile always get the best mileage driving between the two towns on the flat road versus the hilly one..? The question two roads (between two towns A, B) with a different profile cannot be of same length The flat road (1) is shorter, ...

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I believe, the answer is a small but quantifiable, yes, there is a non flat road configuration that would lead to better gas mileage between any two points at the same height. I have numerically solved for such an optimal path. I believe I can give a nice explanation of why that is, but it will take some work, so bear with me. Granted, you can only expect ...

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Assuming a perfectly elastic collision, you are heading in the right direction here. It looks like you're trying to say the thing your ball hits will have 2 times the ball's momentum. You might want to consider $2m_bv_bv'_w=m_w{v'_w}^{2}$ because it looks like something might've gone "wonky" when you added equation (1) and (2)

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Kane's Method is another accepted formalism (Thomas R. Kane) which is a method for formulating equations of motion.

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Answer #11: You need 1. downhill slope to get the car to an optimal speed - corresponding the highest gear+optimal rpm. 2. At the optimal performance regime at the point of upslope start, you start the motor and keep adding the energy necessary to get to B with $v=0$, then stop the motor and get to B. Reasons: I assume there is lower efficiency at lower ...

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Let's review the linearisation and go to the further details. Just the pressure might be not enough. Take the momentum equation: $$-\frac{1}{\rho}\nabla p = \frac{\partial \vec{v}}{\partial t}+\vec{v}\cdot\nabla\vec{v}$$ Here we have to eliminate the convective part $\vec{v}\cdot\nabla\vec{v}$. Usually the argumentation is that changes of the velocity ...

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I will choose in function of the wind factor. Wind is a positive source of energy if it blows from the rear and... The road thru the hills has the potential to minimize the exposure to the wind if it comes from a frontal direction. In the valley the exposure is minimal. The plain road has the potential to maximize the exposure to the wind if it comes from ...

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By adding the rubber band you did two things: increased the air drag unbalanced the fan The bearings of a fan don't like to be unbalanced - the friction goes up significantly because as the fan picks up speed there will be a large lateral force (centripetal force keeping the rubber band plus object in their circular orbit). When you balance the fan it ...

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The load increased but the input power driving the fan remained same. Moreover in very accurate measurements, the air drag can also not be neglected, all this will hold for a very another reason that the blades rotated by the motor are of very much comparable mass to the taped rubber & stuff.

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I'm not familiar with "Modern Analytical Mechanics" by Pellegrini & Cooper so I can't comment on that one but I'm very familiar with the other two books you mentioned. Landau's books are generally excellent but tend to be shorter in length and sometimes very dense. Nearly every paragraph has some profound insight that you'll miss if you don't ponder ...

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These problems are best solved with the Euler-Lagrange equations. Work in polar coordinates centred about the circle. Then the ring is at $(R,\theta(t))$. Its kinetic energy is, $$T = \frac{1}{2} m R^2 {\dot{\theta}}^2$$ From the spring there is potential energy. Let $A$ be $(L,\phi)$ in polar coordinates. The potential energy is proportional to the ...

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Yes - but it needs the right sort of hill. A family member, now sadly passed away, used to conserve fuel during 2nd world war by going into town down a hill with a very long and very gentle incline downwards and turning the engine off - the road is also pretty straight for about 3 or 4 miles. This is hinted at in the answer of Edward. Sorry, this has ...

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Torque $\tau = \vec{r} \times \vec{f} = I \vec{\alpha}$ $r$ here is moment of arm, ie. the distance(& perpendicular distance) from the axis of rotation of the body. You can only have and define torque if you have an axis about which the subject will rotate. It will be an absolute force driven motion, with NO torques compelled to induce.

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