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0

You should think of the definite integral operation as a function of two arguments: a region over which to integrate (here, $[x_0,x_1]$), and another function $f$ called the integrand (here, $f:\xi \mapsto (E-V(\xi))^{-\frac{1}{2}}$). So first of all, in my definition of $f$ above, we could have used (almost) any other symbol instead of $\xi$ and the ...


0

The first equation has just explicited the definition of momentum. As a matter of fact, $p=\partial \mathscr{L} / \partial \dot{q}$


1

The rule is simple: regardless of which forces are acting, if the motion is accelerated then there is a net force, otherwise, there not net force. The only kind of non-accelerated motion is motion in a straight line at uniform speed. In particular you options: a)wrong, at it is accelerated in the curved part. We do not know in the straight part. b)wrong, ...


4

"The number of degrees of freedom can be defined as the MINIMUM number of independent coordinates that can specify the position of the system completely" (wikipedia) In your case the number is ONE, because you only need to know the position of the particle along the curve. It doesn't matter if the curve is not a line, or even contained on a plane, because ...


1

As the counter example given by Herr_Mitesh shows it is not true and this is because the lagrangian is not uniquely determined. In physics sometimes you don't have to think like in mathematics and in this case you must content yourself thinking that if the lagrangian does not contain x as a variable that is enough for the condition of homogeneity to be ...


0

You can calculate the maximum height above the launch point with $$y_c = \frac{v_y^2}{2 g}$$ and draw a horizontal line at this height. Now you must know the initial direction of travel you can fit a parabola to this slope while being tangent to the height line at $$x_c=\frac{v_x v_y}{g}$$. The general shape of the curve is $$y = y_c - K (x-x_c)^2 $$ By ...


-1

It is because there aren't any forces acting on the mud keeping turning with the tire that it flies off. At whatever point the mud comes off, it will travel tangent to the tire at first and the follow a parabola due to earths gravity. It is most likely the more loose mud will come off first and at that point the tangential direction of the tire points ...


3

You were probably expected to note that the path of any point of the tire is a cycloid. At the point of contact with the ground, the tire is not moving. As it rises from the ground it is moving faster, with a speed that is $v(1+\sin \theta)$ where $v$ is the bike speed and $\theta$ is measured counterclockwise from horizontal. The centrifugal force is ...


1

In a classical context, LRL vector is conserved only for potentials behaving like $\frac{k}{r}$, indeed we can see the general construction of LRL vector : \begin{eqnarray} \frac{d\vec{p}}{dt}\times \vec{L} &=& -\partial_r v(r) \frac{\vec{r}}{r} \times \vec{L}, \nonumber\\ \mu r^3 \frac{d\hat{r}}{dt} &=& ...


1

Perhaps I could share some idea for further research. If we could make actual and correct pressure measurements in the cochlea to reveal wether the non-stationary Bernoulli effect is a good description of the actual physics-of-how-the-cochlea-isolates-frequencies-along-its-length? I would consider: I would propose to use a pitot tube, with sensor in the ...


4

It helps to write the full action: $$S = \int \frac{-mc^2}{\gamma}dt - \int U dt $$ The first term can be put in a much better form by noting that $d\tau = \frac{dt}{\gamma}$ represents the proper time for the particle. The action is then: $$S = -mc^2\int d\tau - \int U dt$$ The first term is Lorentz invariant, being only the distance between two points ...


1

The easiest way to see the equality is to use a more general formula for the magnetic dipole moment of a particle. For a flat planar loop of current, it's true that $\mu = IA$, with the direction of the dipole normal to the loop. However, the more general case is that of a a volume current $\vec{J}$ in some finite region of space. In this case, the the ...


0

The direction of the magnetic moment is perpendicular to the plane of the loop. Seeing that the angular momentum is also perpendicular to that plane, and having shown that their magnitudes are proportional, is all it takes to show that two vectors are proportional. If you insist, we can still formally go through all that. One way of enforcing an area vector ...


3

An important connection to relativity can be made here. Consider the infinitesimal displacement in the Cartesian coordinates: $$ ds^2=dx^2+dy^2+dz^2=dx^a g_{ab}dx^b $$ where $a,b\in\{1,2,3\}$ and $$ dx^a=\left(\begin{array}{c}dx\\dy\\dz\end{array}\right) $$ and $g_{ab}$ the metric, $$ ...


4

The cochlea has a complex physical structure, with multiple membranes and fluid-filled chambers. Therefore to explain the separation of frequencies along the basilar membrane of the cochlea is complex to. Sure, there are a lot of very general descriptions (even the answer of theblackcat) and a lot never go into the actual physics of the system. This ...


4

When you find the total (squared) value of some vector in an orthogonal basis, such as the Cartesian system $(x,y,z)$ or indeed the spherical system $(r,\theta,\phi)$, what you're doing is simply adding the squared values of each component of the vector. Taking the velocity, let's think about the different components: What is the velocity in the radial ...


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There is an effortless way, if you accept geometrical reasoning. You know, that $T = \frac 1 2 m \vec v^2 = \frac 1 2 m \lvert \vec v \rvert^2$. Furthermore, spherical coordinates are orthogonal, therefore you can just write: $$\lvert \vec v \rvert = \sqrt{v_\phi^2 + v_\theta^2 + v_r^2}$$ Geometrically, one easily finds: $v_r = \dot r$, $v_\theta = r \dot ...


3

What are the main differences between a quantum and classical system? How does one can distinguish them? An experimentalist's answer, in other words, why do we need two different theories. In a classical system if we do experiments with massive bodies , they have in principle well measured (x,y,z) positions at time t. From falling apples to the ...


-3

In a classical system (as we have studied), the behavior of a body can be exactly determined. For example, in projectile motion, the position, velocity, and acceleration of a system can be determined exactly by knowing only the time $t$ plus your initial conditions. In terms of size a classical system usually deals with macroscopic systems. In a quantum ...


1

The main difference between classical and quantum physics is the fact that observables (Hermitian operators whose eigenvalues determine the possible values of physical quantities of the system) do not commute. For classical systems commutativity is trivially satisfied. A direct consequence of this is the uncertainty principle and the fact that via Bell's ...


-1

$\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$ $\newcommand{\ket}[1]{\left| #1 \right>}$ Quanta vs. Continuous Quantum, as the name suggests, deals with quanta ie packages of something. For example for a given particle and given potential the energy is quantised. Eg. for quantum mechanical SHO the energy levels are $E_n=\hbar \omega ...


5

It's risky to think about subatomic particles in a classical way, but maybe we can get something from it if we're careful. Electrons orbiting an atom in a state with well-defined angular momentum quantum number $\ell$ have wavefunctions described by the spherical harmonics. The $s$ shell, with $\ell=0$, has spherical symmetry; this state really is ...


2

$\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}$ $\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$ $\newcommand{\l}[0]{\mathcal L}$ $\newcommand{\q}[0]{\dot q}$ $\newcommand{\pdvt}[2]{\frac{\partial^2 #1}{\partial #2^2}}$ A word of advice: Landau's book, don't read it if you haven't studied that subject before! Your first question has ...


3

If we think as electrons around atoms classically, then electrons would irradiate electromagnetic energy, losing momentum and thus collapsing into the nucleus, and atoms couldn't exist. Therefore your question makes no sense. The correct description of an atom is using quantum mechanics, which means there is no orbits on atoms. There is only the solution of ...


1

You apply the chain rule basically, so I'll try help you with the first sum and leave the second for you. http://imgur.com/a/QyYtx your final comment is correct but the first sum should be over a dummy variable j, and the second should summed over dummy variables j,k. FrolovOut


4

The part at the exact middle of the string has zero mass. That seems silly, but consider - if you consider a very small section of the string in the middle - say 1 mm - then the pieces of string on either side exert forces with tiny, but nonzero upward components. If you half the length we are considering to 0.5 mm, then the upward component of the forces ...


7

Note that this is almost identical to this previous SE question; the answers on that question however, including the favourite answer with 13 upvotes, are somewhat erroneous (or rather oversimplified). This is a good question: the exact details of how such elevations, called 'central uplifts', form are not well known. What follows is a brief summary of ...


1

Consider a volume element $dV$ at a certain point $\vec{x}$. Let the strain tensor at it be given by \begin{equation} e_{ij} = \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \end{equation} Let us diagonalize the strain tensor at this point and let its diagonal entries be $u^{(i)}$. Since the trace of tensor is invariant, ...


2

how on earth.....? A possible way to look at it goes like this, let's consider a litle cube then the stress force in the ith direction acting on jth surface element is $-\sigma_{ji}(x_i,x_j,x_k)*dA$ where $dA=l^2$, the force in the ith derection acting on the other surface element paralell to the first is $\sigma_{ji}(x_i+dx_i,x_j,x_k)*dA$, so the total ...


0

Terminal velocity of a free falling object is obtained at the moment its acceleration vanishes \begin{equation} \Sigma\mathbf{F}=0. \end{equation} The forces that act on the object while falling are the gravity force, \begin{equation} F_g=-mg, \end{equation} and the drag force \begin{equation} F_d=\frac{1}{2}v_t^2dC_dA \end{equation} where $v_t$ is the ...


1

is my interpretation of the dynamics of the self-force correct and is there a physical or intuitive explanation for this extremely pathological behavior in the presence of a Coulomb potential? Eliezer makes his argument based on the equation with the Lorentz-Abraham-Dirac term. This term was originally (Lorentz) devised as an approximate way to account ...


3

There is a rotational equivalence, but it is not what you stated. The problem, as pointed out by @curiousOne, is that conservation of angular momentum does NOT imply rotation about the same (fixed) axis. But I think a simple restatement like this could work: if no torque acts on a body, its angular momentum will remain unchanged rate of change of angular ...


2

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 ...


0

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 ...


1

For our three compartment hearing sense, from a physics point of view, there is a basilar membrane stimulation, from base to apex, in its pathway in the cochlea, to a place on the basilar membrane. By periodic movement of perilymph, non viscous fluid, backwards and forewards, in the cochlear duct meet the conditions of a potential flow. The basilar ...


1

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 ...


1

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 ...


2

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 ...


2

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 $\frac{d q^k}{dt}$ ...


0

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 ...


0

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.


3

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: $$ ...


3

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 ...


1

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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