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# Tag Info

## Hot answers tagged classical-mechanics

4

The dogs are definitely doing more work when chasing the cat over a given distance. The distance is the same but the force is larger : your motion is not accelerated, which means that the net force on you is zero. The dogs are pulling you through their leash, and your feet are resisting this force by friction, except for an extremely small resultant to ...

4

Granted appropriate boundary conditions, the stationary action principle and the Euler-Lagrange (EL) equations are both precisely the condition that the functional/variational derivative $$\frac{\delta S}{\delta x^j (t)} \tag{1}$$ vanishes, so they better agree!

3

First, I think there is something wrong with your partial derivative of the Lagrangian with respect to $x$. Second, the Euler-Lagrange equations are nothing more than the process that you performed in Method 1, done without committing to a specific form for $L$ but leaving it generic. In your first step you took partial derivatives of $L$ with respect to ...

3

Take a lift from the first floor to the tenth floor. You feel getting heavier when the lift leaves the first floor and getting lighter when it arrives at the tenth floor. You can explain this by noticing that the lift is not an inertial frame. It's accelerated and later slowed down, relative to earth (which is not strictly speaking an inertial frame, but ...

2

The dog is doing more work becuse work = (force)(distance).So your dogs pull harder on you (more force) but don't move any faster so force is more while distance remains the same.

2

If the dogs pull harder they do more work. Trivially: work is force times distance. The wording of your question leaves a small crack of doubt as to whether they are actually pulling harder after they see the cat: you include "(apparent)" in the title, and in the question say "the result was two dogs frantically trying to pull" [emphasis added]. Turns out,...

2

Assuming from the notation $$\dot{x}^i~=~f^i(x,p,t), \qquad \dot{p}_i~=~g_j(x,p,t), \tag{1}$$ that the symplectic structure is the standard canonical symplectic structure $$\omega = \sum_{i=1}^n\mathrm{d}p_i\wedge \mathrm{d}x^i,\tag{2}$$ we get that \begin{align}\mathrm{d}H(x,p,t)- \frac{\partial H(x,p,t)}{\partial t}\mathrm{d}t ~=~&\sum_{i=1}^n\... 2 The big O term is ignorable because we are showing that a necessary condition for closed orbits is that the potential harmonic or inverse square. We are led to investigate the stability of circular arbits under that perturbations are arbitrarily small so the O term is arbitrarily small and can be ignored. If we going the other way --- trying to ... 2 No, this is not always true. According to conservation of energy, the change in total mechanical energy of a system is determined by the work done by forces external to the system:\Delta E=\Delta K+\Delta U=W_\text{ext}$$What you are proposing is true if W_\text{ext}=0, because then as \Delta K\geq0 it must be that \Delta U\leq0. Since we started ... 2 The nice thing about the delta distribution, and its derivatives, is they only act at a point. Let's consider the trajectory of a particle which is initially not located in the potential (i.e. not at q_0). In this case, the motion is as for a free particle, at least initially. In fact, if we can conclude that the trajectory of the particle never approaches ... 1 Since F_{Net}=ma, if m=0 for pulley Q, then you can only have non zero acceleration of the pulley if F_{Net}=0 also. If we can assume the string is ideal (massless and inextensible), and there is no friction between the string and the pulleys, the tension, T, throughout the string must be the same. (I said assume because the complete problem ... 1 Think of one particle of a layer. It is moving in simple harmonic motion around the center, since$$a\propto x$$Now think of the whole layer. The whole layer is moving together. If you can analyze how one particle of this layer moves than you know how the whole layer moves. If you understand how each layer moves then you understand how the whole sphere ... 1 Well, given$$ \dot{p} = - \frac{\partial H}{\partial x} \quad \text{and} \quad \dot{q} = \frac{\partial H}{\partial p} $$we have$$ H = - \int dx_{i} \, g_{i}(x,p,t) $$and$$ H = \int dp_i \, f_{i} (x, p, t). $$You deal with the constants of integration using the obvious constraint:$$ - \int dx_{i} \, g_{i} (x, p, t) = \int d p_{i} \, f_{i} (x,...

1

They may in special cases, but this is by no means a rule, and most certainly doesn't hold for all cases. As I'm a firm believer in concrete examples, I'll illustrate this with the following simple case of two blocks, each of mass $m_1=m_2=m$, connected to walls and each other by Hookean springs (which are harmonic oscillators) of spring constant $k$, as ...

1

You are nearly finished. Consider your result $$L' = L+ \theta(2xy(a-b)-3cxy^2) + O(\theta^2).$$ In order to make $$\theta(2xy(a−b)−3cx^2) = 0 \quad \text{for all } x \text{ and } y$$ you must have the conditions $a=b$ and $c=0$.

1

@Jame's answer is a more simplistic one to approach the problem. However, I agree with him that the dogs will be doing more work compared to if they don't pull the leash. But I want to discuss a case here of by "how much", it is not directly linear after all. Again, if we review, work is a dot product of two vectors. As a general rule, in your example, the ...

1

This is a perennial question on this site, and always leads to confusion over the distinction between "work" and "effort". To avoid confusion, we need to be explicit about the many terms in the energy. Energy contributions Crudely, the energy of the dog can be written as the sum of four terms, E = U_{\text{pot}} + K_{\text{KE}} + U_{\text{chem}} + K_{\...

1

My previous answer with the example of the dog exerting the same force but doing less and less work I believe proved that physical effort does not always equal work as defined by Physics. That is how I interpreted, perhaps incorrectly, what your question was really about. But while it is also true, as @James points out, that if the dogs exert a greater ...

1

Work in physics has a specific meaning. Force times distance over time. Imagine that your leash is connected to a 100 pound weight which is on ice skates, that travels on ice. You however have an ice-free concrete sidewalk to walk on. Imagine that you pull with a constant force. The weight starts to move slowly. As you add force it speeds up, and then it ...

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