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1

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 ...


2

It's the response of the system to a stimulation at zero frequency. In other words, it tells you the displacement of the system in equilibrium under a time independent force. Let me give an example. Consider a mass on a spring with friction and an external force $F_{\text{ext}}(t)$. The friction force is $$F_{\text{friction}} = -\mu \dot{x}$$ so the ...


1

The fastest way is to compare kinetic energies in the two cases: \begin{align*} KE &= \tfrac{1}{2}I_{\text{cm}}\omega^2_{\text{cm}} + \tfrac{1}{2}M(R\omega)^2_{\text{cm}} \\ KE &=\tfrac{1}{2}I_{\text{inst}}\omega_{\text{inst}}^2 = \tfrac{1}{2} (I_{\text{cm}} + MR^2)\omega^2_{\text{inst}} \end{align*} So $\omega_{\text{inst}}=\omega_{\text{cm}}$. The ...


0

which should not be possible. Indeed, uniform coordinate acceleration $a$ is inconsistent with special relativity however, uniform proper acceleration $\alpha$ is consistent. The proper acceleration is the acceleration of the object according to an attached accelerometer. For 1D motion, the relationship between $\alpha$ and $a$ is given by $$\alpha = ...


14

Have a look at the article by Phil Gibbs on the relativistic rocket. This describes the motion of a rocket that is accelerating with a constant acceleration. In this context constant acceleration means the crew of the rocket feel a constant acceleration. Technically the rocket has a constant four-acceleration. Anyhow, the velocity of the rocket as observed ...


1

Look at sparknotes.com/physics/specialrelativity/dynamics/…, you can see $dE/dx=F$ - if your force is constant, it is the energy that increases constantly. $E=\gamma(v)m_0c^2$, you can deduce the $v$. Beacause of laziness I used mathomatic, and it gives me something like this: $v=c\sqrt{1-\frac{m_0 c^2}{(F\cdot x + m_0 c^2)^2}}$ If you check it for x=0 and ...


6

Is there some other formula ... which ... does not allow the speed ... to surpass the speed of light? That would be the equations of special relativity mentioned by sahin in a comment. Image from Loodog? Another factor you have to take into account with classical mechanics is to work out how a constant force can be applied to your object over 11 ...


0

The angle between the object being launched and the platform from which it is launched should remain the same to all observers unless they are moving at speeds comparable to the speed of light relative to the platform. At these speeds they will observe length contraction, the length along the axis which the objects are moving relative to each other will ...


0

$$\dfrac{\partial\dot{r}_i^T\dot{r}_i}{\partial\dot{q}_j}=\dfrac{\partial\dot{r}_i^T\dot{r}_i}{\partial\dot{r}_i}\dfrac{\partial\dot{r}_i}{\partial\dot{q}_j}=2\dot{r}_i^T\dfrac{\partial\dot{r}_i}{\partial\dot{q}_j}$$ (A simple chain rule, while bearing in mind that $\dot{r}_i^T\dot{r}_i$ is essentially $\dot{r}_i^2$. You just need a transposed vector on the ...


2

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.


3

The Galilean spacetime is indeed the affine space $\mathbb{A}^4$. Affine space can be considered as a 'space with no origin', which makes intuitively sense because why would some point (the origin) be special. For example a trivial Galilean space is $\mathbb{E}\times \mathbb{E}^3$ where $\mathbb{E}$ is Euclidean space. The $\mathbb{R}\times \mathbb{R}^3$ ...


3

The only problem with your hypothetical ultralight black hole is that it would be big. Very big. First, note that if you had a black hole with mass $M$ and corresponding volume $V$ such that $M/V < \rho_\mathrm{space}$, then a volume $V$ of a typical patch of space would have more than enough mass to be a black hole. This should worry you if you thought ...


-4

Because of hp and torque. When in a low gear, the rpms are higher and the hp and torque kicks in at higher rpms.


2

I'll do all calculations assuming the lagrangian $\mathcal{L}$ acts on a 1-dimensional manifold $M$. I believe you'll find the generalization absolutely trivial, and this will spare me of writing tons of sums. Let \begin{equation} \mathcal{L}: \mathbb{R} \times T M \rightarrow \mathbb{R} \end{equation} be a lagrangian over $T M$, with time in ...


0

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 ...


0

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 ...


0

This is a problem that involves only calculation of velocities from other velocities, no influence (forces) needs to be considered. Imagine the system as it appears in the inertial system where the point of contact of the two wheels is at rest. Since the contact point is at rest, the mass points of both wheels that touch each other are at rest. Since any ...


1

The columns of a 3×3 rotation matrix contain the coordinates of the local xyz axes (expressed in world coordinates). With Euler angles (321) you apply the elementary rotations $R_Z$, $R_Y$, $R_X$ in sequence to form the local → world rotation matrix. That is $$ E = R_Z(\varphi) R_Y(\psi) R_X(\theta) $$ This is interpreted as each rotation occuring about ...


2

Center of mass before the disintegration is in the initial particle. This means that the center of mass moves with velocity $\vec V$ in lab frame. Thus, to switch from center of mass frame to lab frame you just use the Galilean transformation $$\vec v=\vec v_0+\vec V.$$ As for conservation of mass, it is certainly implied because the system is considered ...


1

Let's say you do the calculation $C_{final}=R_xR_yR_zC_{o}$, where the $C$s are coordinates. The first rotation is about the $z$ axis of $C_{o}$ and will produce a new coordinate system in which the new $z$ is the same as the old, but the $x$ and $y$ axes are different. The next rotation will be about the new $y$ axis and will produce a newer, new $x$ and a ...


1

We have such equation: $$H = \frac{\partial L}{\partial \dot{q}} \dot{q} - L$$ You can show by calculation, that it holds in your special case, too. Now we use chain rule, and Euler-Lagrange equation: $$\frac{dL}{dt} = \frac{\partial L}{\partial q} \dot{q} + \frac{\partial L}{\partial \dot{q}} \ddot{q} =\frac{d}{dt}\left(\frac{\partial L}{\partial ...


6

Comments to the question (v2): To go from the Lagrangian to the Hamiltonian formalism, one should perform a (possible singular) Legendre transformation. Traditionally this is done via the Dirac-Bergmann recipe/cookbook, see e.g. Refs. 1-2. Note in particular, that the constraint $f$ may generate a secondary constraint $$g ~:=~ \{f,H^{\prime}\}_{PB} ...


1

The Hamiltonian is defined by $$ H = \sum_{i=1}^n \left( \frac{\partial L}{\partial \dot q_i} \dot q_i \right) - L $$ So in your case: $ H' = H - \lambda f $


2

Quantum mechanics, excluding gravity, is valid at all scales. In principle, you can do classical mechanics via quantum mechanics, but this is just very hard because of the number of particles involved and decoherence. Therefore, you switch to classical mechanics, which describes an effective theory of large scales (and small energies), because most of the ...


1

This quite special top is called a rattleback, or celt. See Wikipedia : http://en.wikipedia.org/wiki/Rattleback I quote : "The spin-reversal motion follows from the growth of instabilities on the other rotation axes, that are rolling (on the main axis) and pitching (on the crosswise axis). (...) The amplified mode will differ depending on the spin ...


1

While I cannot possibly claim to know what Feynman did, here is a fairly-simple way of deriving the 2:1 ratio (which incidentally Feynman remembered backwards in that quote; the wobble is faster than the spin). The viewpoint I'll take is that it's a given that the spin rate and wobble rate are constant, then ask why they are related. We'll work with a thin ...


0

At the moment you are on the physics part of this site. As I'm sure you know, some computer languages are written especially for solving maths problems and nothing else. If you also posted your question on this site http://scicomp.stackexchange.com/ I think you will find someone who can help you. There is probably code already written that might solve ...


2

The simplest approach (and what the person asking this question probably was getting at) is to use a pulley like so: The weight of the object is now shared between the two sides, with each carrying a 50 N load. You end up using twice as much cord. The other advantage is that you now have a "mechanical advantage" and you only need to use a force of 50 N to ...


1

I suspect this is possible in the same way as when you strike a billiard ball below its center of mass: the ball moves forward, stops, and moves back because when it stops, it still has rotation energy and tries to keep rotating, but friction opposes this rotation and causes the ball's motion back. So the plastic piece probably still has some rotation ...


2

They do make such flywheels. From Wikipedia: "Some modern flywheels are made of carbon fiber materials and employ magnetic bearings, enabling them to revolve at speeds up to 60,000 RPM."


1

At least three different quantities in physics are customary called an action and denoted with the letter $S$: The off-shell action functional $S[q;t_i,t_f]$, The (Dirichlet) on-shell action function $S(q_f,t_f;q_i,t_i)$, and Hamilton's principal function$^1$ $S(q,\alpha, t).$ For their definitions and how they are related, see e.g. my Phys.SE answer ...


0

1) Well, constrained Hamiltonian dynamics is a huge subject. Quoting Wikipedia: A dynamical quantity is a 1st class constraint is if its Poisson bracket with all the other constraints vanishes on the constraint surface (the surface implicitly defined by the simultaneous vanishing of all the constraints). A 2nd class constraint is one that is not ...


1

Second possibility: $$\dfrac{\partial}{\partial u}\int u''^2 \mathrm{d}x=\dfrac{\partial}{\partial u}\int > u\,u^{(4)}\mathrm{d}x=\int u^{(4)}\mathrm{d}x$$ by double integration by part and because $\dfrac{\partial > u^{(4)}}{\partial u}=0$. I am really not sure about this latter argument either. Second possibility is closest to correct. ...


4

Yes, it holds referring to the Heisenberg evolution of operators and in the specific case of the harmonic oscillator: $$\hat{x}(t) := U(t)^\dagger \hat{x} U(t)$$ where $U(t) := e^{-it\hat{H}}$ (here $\hbar:=1$). One has $$\frac{d^2}{dt^2} \hat{x}(t) =\frac{d}{dt} \frac{d}{dt} U(t)^\dagger \hat{x} U(t) = \frac{d}{dt} U(t)^\dagger i [\hat{H}, \hat{x}] U(t) = ...


3

By Ehrenfest's theorem you have $$i\hbar\frac{\text d}{\text dt}E_\omega[q] = E_\omega[[q,H]]$$ and $$i\hbar\frac{\text d}{\text dt}E_\omega[p] = E_\omega[[p,H]]$$ where $E_\omega$ indicates the expectation value over the state $\omega$. Simple computations show that the first equation gives $$i\hbar\frac{\text d}{\text dt}E_\omega[q] = \frac{i\hbar}m ...


1

If you built a surface such that its height was proportional to a horizontal coordinate $x^2$, $h = k x^2$, then the potential energy at point x would be $mgh = m g k x^2$, which is a harmonic potential. In other words: yes, the curve exists and it's a parabola. This assumes uniform $g$, though, I guess if you want to be a stickler you could note that if the ...


0

When a ball rolls down a ramp, it will accelerate more slowly than a sliding object (without friction). This suggests that you could treat the motion as that of a sliding object with a larger apparent inertial mass. I believe you will be able to solve your problem if you take this hint and apply it to your situation. You might find some inspiration in this ...


2

I invite you to read the following papers about "Lie group Thermodynamics" of Jean-Marie Souriau. Souriau has discovered that Gibbs equilibrium is not covariant with respect to Dynamical groups, then he has considered Gibbs equilibrium on a Symplectic Manifold with covariant model with respect to a Lie group action. Souriau has introduced a geometric ...


0

It is proportional to its cross-sectional area. "Muscle strength depends on the number of fibers in a particular muscle. Hence the strength of a muscle is proportional to its cross-sectional area" source: Conceptual Physics by Paul G. Hewitt


3

The equations you wrote down describe a wave that is traveling one way, then gets reflected and travels back the other way. The leading edge of the returning wave changes the slope of the wave - just before it becomes a full standing wave it is a mixture of the traveling wave (to the right) and the leading edge of the reflected wave. At that leading edge, ...


0

I believe it means that there is a kink or cusp in the pattern the wave makes at that point in time, presumably due to the fact that there was some external disturbance needed to make that change happen.


0

Remember that in a silenoid work is done because something moves. The motion gives rise to a change in flux which results on an e.m.f. - and the product of this and the current flowing is the work done. For the same motion, there will be a larger emf when the inductance of the coil is greater. Greater emf times same current = more work done. Put ...


0

You made a mistake. Let me explain it simply by taking two waves that begin simultaneously from opposite sides of the chord. The waves won't look as you say, but $$f_{+}(x, t) = Asin(kx - \omega t) \tag{i}$$ and $$f\_(x, t) = Asin(-kx - \omega t) \tag{ii}$$ i.e. they advance toward one another from opposite direction. Their superposition produces ...


0

I've realized I made a very silly error in calculation (despite thinking about this for a little too long...) $y(\tau/2)$ doesn't vary with time, so obviously $\frac d{dt} y(\tau/2)=0$, since $y(\tau/2)$ is a constant value with respect to $t$.


0

Let us consider this with two symmetrical pulses traveling in opposite directions; they are exactly alike except that one is positive & other negative. As they pass through each other, there comes a moment at which the whole string is straight. Where did the energy go during the annihilation? After some time, the pulses reappear. But "what is it that ...


0

For question 2 it indeed makes sense to reuse the approximate solution since you consider that the real solution is not very different from the one you have already calculated. I think that when you write : $-(g / \ell) \theta (1-\theta ^2 /3!)\approx -(g / \ell) \theta (1-\frac{<\theta ^2 >}{3!})$ you are making an approximation at 2 levels : first ...


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 ...


0

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 ...



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