# Tag Info

21

It is quite simple. It is that you are entirely overlooking the fact that the marble is interacting with the track and with whatever is supporting the track and ultimately with the Earth. The combined momentum of the the marble, the track and the Earth is conserved throughout. If you repeated the experiment with the marble on a light track which was on a ...

9

Well, it seems to me that under a translation $x(t) \to x(t) + c$, the Lagrangian goes to $$\mathcal{L} \to \mathcal{L}' = \frac{1}{2}m v^2 - mg(x+c) = \mathcal{L} -mgc.$$ So yes, the Lagrangian may appear to be different, however since it only shifts by a constant, these two Lagrangians ($\mathcal{L}$ and $\mathcal{L}'$) are equivalent and produce the same ...

7

The momentum of a particle running along a track is definitely not conserved. Momentum is only conserved when there are no external forces applied to a system, and the track here applies forces to the marble. What is conserved for a frictionless track is mechanical energy- the sum of kinetic and gravitational potential energy. But while you can spin the ...

7

Well, assuming the resulting force is non-zero and F=ma holds, $$\lim_{m\to0} ma = F,$$ which would lead to an infinite acceleration. But that would not contradict the intuitive observations per se - the smaller the object, the faster it could plop away from these two massive colliding bodies. Unsurprisingly, this "proof" does not really prove ...

6

You can consider the addition of mass to be an inelastic collision where kinetic energy is not conserved by the momentum is conserved. The linear momentum in the horizontal direction of motion should remain the same before and after the collision. $$mu = (m+M)v$$ where $m$ is the mass of the trolley, $M$ is the added mass, u is the initial velocity of the ...

5

The third law is not valid for 2 charged particles moving around as result of their interaction. The combined electric and magnetic forces are in general different in modulus and don't have the same direction. Of course, the momentum is conserved when taken in consideration the fields. But it is beyond that type of proofs.

4

Here are a few observations, that I think together form an answer to your question (although I'm not sure I followed all of the notation in the body of the question). Any solution to a (scalar) second order differential equation has two constants associated with the solution, namely the two integration constants. However, in general, we usually don't call ...

3

Momentum is always conserved. Perhaps you are thinking of a system where the mass is brought up to speed with the trolley, then gently attached to it? In that case, then the trolley will, indeed, have more momentum than before the addition -- precisely as much as the mass had by itself! If the trolley crashes into the mass, then you're talking about the &...

3

Your system is the mass and trolley, so that before the mass is added, the total momentum is that of the trolley only since the mass is presumably not moving. Now when the mass is added (gently so that there is no downward velocity component), the trolley will move with a different velocity, but the total momentum of the trolley and mass system will be ...

3

Think about the energy for a particle of mass $m$ in a uniform gravitational field, $$E=\frac{1}{2}mv^{2}+mgz.$$ This energy is conserved, because it does not depend explicitly on $t$. That essentially just means that $t$ does not appear in the mathematical expression. However, if the mass were changing (if, say, it was being eroded by wind), making the ...

3

Energy conservation means that the equations of motion have a first integral. This is a non-trivial statement, since it selects a subset of equations out of all imaginable types of them. While many equations of motion obtained empiriclaly were known to have such an integral, the Noether's theorem makes this statement in the most general form, as a result of ...

3

You can do an integration by parts on the last term (and discard the resulting boundary term) to yield an action with equivalent EOMs:$$S'= \int \left( \frac{1}{2} m \dot{x}^2 + m g t \dot{x} \right) \, dt$$ In this context, the symmetry $x \to x + C$ is obvious at the level of the Lagrangian. Moreover, the Euler-Lagrange equations become $$\frac{d}{dt} \... 2 Don't forget about the Earth. In the vertical direction, the system of the masses A and B move down, and the Earth ever-so-slightly moves up. Similarly, in the horizontal direction, because of the rope in the pulley, a force is directed to the table which may be fixed to the Earth. You'll find a similar paradox if you push yourself away from a wall. You'll ... 2 If we assume the whole system is isolated and in equilibrium, then it is in its maximum entropy state. Remember that isolated systems evolve until they reach equilibrium, and irreversible processes, like equilibration, cause the entropy of an isolated system to increase. Once the system reaches equilibrium, it stops evolving, and its entropy remains ... 2 Assumptions I take the static friction is enough that the motion of the marble does not cause the ramp to oscillate on the floor. Let's try to imagine the situation, attach an momentum arrow to the marble as it moves. Let us try deduce facts about this arrow as the marble moves. The time rate change of magnitude of the momentum vector is given as:$$ \frac{...

2

On one hand, Noether's theorem in its original formulation assumes an action formulation. OP's setup (v1) lacks this. The action formulation leads among other things to the standard formula for energy (which is of course the Noether charge for time translations), cf. e.g. this Phys.SE post. On the other hand, given only OP's triple, there is not a clear ...

2

It follows from the principle of conservation of momentum. Think of it as a transfer, like a payment. If you give me 20 dollars, I have 20 dollars more than I had before and you have 20 dollars less because you have transferred money to me. Giving me 20 dollars has an equal and opposite effect on the amount of money you have. It is the same with forces. If ...

2

You are correct that $r$ and $\theta$ are functions of time, and Goldstein is not claiming that they arent. The whole point of the discussion is that $p_{\theta}$ is a constant, even though it may not appear so at first since $r = r(t)$ and $\theta = \theta(t)$. This follows from the Euler Lagrange equations, there is one for each generalized coordinate $$\... 2 To understand the conservation of momentum here, it is necessary to think about the forces involved, because the principle of conservation of momentum has a dependent clause: If the net external force on a system is zero, then the momentum is conserved. Because momenta and forces are vector quantities, this actually holds true for each component of an ... 2 You're right to be suspicious of this proof. It's wrong. The error is most succinctly stated in the fourth paragraph: [for] a body of mass zero, [...] Newton's second law tells us that F_{\rm net} = 0a, so F_{\rm net} = 0. If "bodies" are things that Newton's laws describe then there are no "bodies of mass zero". You can't push a ... 2 Here, \mathscr{d}m is inherently negative because rockets mass decreases with time.$$ \mathscr{d}m = -|dm| $$And the rocket's mass is,$$ m + \mathscr{d}m = m - |\mathscr{d}m| $$For reference, please look Young's University Physics with Modern Physics, Ch 8, Sec 8.6 1 Assuming you are talking about projectile motion in the presence of a uniform gravitational field the answer is no. The total energy of a particle of mass m is given by E=\frac{1}{2}mv^2 +mgy  where y is the height of the projectile from the floor and v^2 is the magnitude of the velocity squared. As the projectile launches up kinetic energy is ... 1 Just a general comment on this type of question. Many are saying "laws cannot be proved". While, in principle, this sounds reasonable it's not really true in practice. Some counterexamples: We find out from the principal of least action that Newton's 2nd Law can be "proved". We find that the "entropy law", or Clausius' theorem, ... 1 Mass ~m_A~ is moving with the initial velocity ~v_{A0}~, mass ~m_B~ is moving with the initial velocity ~v_{B0}~, both collide with mass ~m_C that doesn't move the collision equations are:$$m_{{A}} \left( v_{{A}}-v_{{{\it A0}}} \right) =-{\it dp}_{{A}}\\ m_{{B}} \left( v_{{B}}-v_{{{\it B0}}} \right) ={\it dp}_{{B}}\\ m_{{C}}v_{{C}}={\it dp}_{{A}}-...

1

The supposed 'proof' you quote is entirely specious. The force imposed by each of the larger balls on the smaller would tend to zero as the mass of the smaller ball tended to zero, so you would simply have the result that the overall force was zero because the two contributions to it were, and not because they remained large, equal and opposite. You cannot '...

1

For example: I do a force on a broken car to move it. But instead of put my hand directly on it, I hold a spring, and there is another equal spring fixed on the car. So I do the force by contacting both springs. No matter if the force is enough to move the car or not, the deflections of the springs are the same. So the forces are equal. We could imagine a ...

1

The author Laura Becker of OP's linked webpage is considering 2 types of actions: The off-shell action functional $$I[q;t_i,t_f]~:=~ \int_{t_i}^{t_f}\! {\rm d}t \ L(q(t),\dot{q}(t),t), \tag{1}$$ The Dirichlet on-shell action function $$S(q_f,t_f;q_i,t_i)~:=~I[q_{\rm cl};t_i,t_f], \tag{2}$$ where $q_{\rm cl}:[t_i,t_f] \to \mathbb{R}$ is the extremal/...

1

There are two ways to arrive at energy and momentum in electromagnetic fields. First approach: start from force on charged particles, and insist on energy and momentum conservation over all. Second approach: start from a field Lagrangian density and use Noether's theorem. The second approach is often preferred by theoreticians who are used to this ...

1

The conservation of momentum is used to know the velocity of the pair monkey + acrobat just after they join. The momentum immediatly after the pair is formed must be equal to the momentum just before the acrobat take the monkey. The rest of the problem is solved by uniformly accelerated movement.

1

The (infinitesimal) translation $$\delta x~=~\epsilon$$ changes OP's Lagrangian with a total time-derivative $$\delta L~=~mg \epsilon~=~ \frac{d}{dt}(mg \epsilon t).$$ This is known as a quasi-symmetry. Noether's theorem does also hold for quasi-symmetries. Concerning symmetries of action vs. EOM, see also e.g. this related Phys.SE post.

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