# Tag Info

19

You can have multiple forces exerted on an object that add to zero. Then there will be no momentum change. Think of the two of us leaning against opposite sides of a door with the same force. The door does not change momentum, nor does either of us. I am exerting a force on my chair as I sit here.

14

Actually Newton's second law is better stated as $$F=\frac{dp}{dt}$$ and this is even valid in relativity, both SR and GR, expressed in the right way $$f^\mu = \frac{dp^\mu}{d\tau}=m\frac{du^\mu}{d\tau} = m u^\nu\nabla_\nu u^\mu$$ (for massive particles) so classically forces are always imply a change in momentum. In QFT the concept of a force is no more ...

10

General remarks. The momentum you define in the first equation, namely \begin{align} p = \frac{\partial L}{\partial \dot q} \end{align} is not necessarily the same momentum that appears in Newton's Second Law. This momentum is called the canonical momentum conjugate to $q$, and it can be quite different from the momentum you're used to (the one ...

7

No, all forces involve a change in momentum. In classical mechanics force is defined as a change in momentum. In quantum field theory particles interact via exchanging one or more bosons (see feyman diagrams). These bosons always have momentum and therefore the momentum of the interacting particles changes as well.

4

You're missing a somewhat subtle point in your analysis. The block on the left in your diagram, where the spring is at its equilibrium position, is moving, so it has kinetic energy (which you're currently ignoring). I'll leave it to you to sort out what the speed needs to be and check that CoE holds. It needs to be moving because, if it were not, then there ...

4

The information about the forces is just as important in predicting the time evolution of a classical mechanical system. When people say that the initial position and momenta are "all" that's needed to predict the evolution of the configuration of the system, they do not mean this in the strict sense that you're indicating. In particular, the statement ...

4

Well, if I read the problem correctly then your kinetic energy is wrong. Your $\dot{x}_m$ is the $x$-component of the velocity of the ball, but you're missing the $y$-component and also the velocity of the cart, not to mention you're multiplying $\dot{x}_m^2$ by $m$ when you should be multiplying it by $M$. The $\cos^2\phi$ goes away because the ...

3

Why don't you just write the equations of motion? You will find that you get a second order lienar homogeneous DE with constant coefficients. Such an equation always has a closed form solution in terms of exponentials, so you can solve it; I guarantee the solution is not too bad. You might even recognize the equation without having to solve it. Edit: OK, ...

3

It depends on how you "derive" Lagrange's equations, whether taking Newton's laws as fundamental or by assuming an action integral and minimizing it. However, there is no such requirement that you be in an inertial frame of reference. Thus, to look at your pendulum problem, you could start with the Lagrangian L = \frac{1}{2} I ...

3

I) In e.g. Ref. 1 is shown that there exist (possibly velocity-dependent) generalized potentials for all the fictitious forces, such as, e.g., the centrifugal force, the Coriolis force and the Euler force. So Yes, there exist Lagrangian formulations for non-inertial accelerated reference frames. II) OP's image shows Kapitza's pendulum. Kapitza's pendulum ...

3

Let's compare Landau's Lagrangian and the one given by $\mathcal L_{Me}$ in your question: $$\mathcal{L}_{Landau}-\mathcal{L}_{Me}=mal\gamma^2\sin(\phi-\gamma t)-alm\gamma\dot\phi\sin(\phi-\gamma t)=\\ =mal\gamma\left((\gamma-\dot\phi)\sin(\phi-\gamma t)\right)=\\ =mal\gamma \frac{\text{d}}{\text{d}t}\cos(\phi-\gamma t)$$ Now the difference is obviously a ...

3

An approach alternative to that discussed by David Bar Moshe is to start from a different coordinate system in the Rindler wedge $W_R$: $$ds^2 = e^{2y}(−g^2dt^2+dy^2)$$ here $t, y \in \mathbb R$. The relation with the standard spatial coordinate in $W_R$ is $x=e^y$, where $x>0$ is related with the alternate form of the (same) metric: $$ds^2 = -g^2 x^2 ... 3 Many clouds are sustained by upward currents, either thermals or generated by a front, that also determine their vertical extension. However this would be an incomplete answer. Look at the clouds as regions where temperature and pressure are such that water molecules can condensate. If a water drop leaves that region without being big enough, it just ... 3 Regardless of whether the "local" situation is symmetric or not with respect to loosening and tightening, what you essentially have is a random walk. At any point in time, the screw can stay where it is, get a little looser or get a little tighter. There is, in practical terms, a limit as to how tight the screw can get but no limit on how loose. For any ... 3 The "missing" energy you're referring to actually left the block-spring system when the external force was interacting with the block. One way to think about it is the following. The work-kinetic energy theorem tells us that, since the KE of the block doesn't change during the lowering process, the net work done on the block is 0:$$W_{net}=\Delta ...

2

Hamilton was guided by a hunch that since a minimum principle worked for optics, then perhaps a similar principle worked for mechanics. From his principle of least action he was then able to derive the Euler-Lagrange equations from his paper: W.R. Hamilton, "On a General Method in Dynamics.", Philosophical Transaction of the Royal Society, 1834

2

The dynamics of a classical point particle moving in the background of any curved space-time is always Hamiltonian (with respect to the canonical symplectic form), thus automatically satisfying the Liouville’s theorem. This is because the action functional is given by the integral of the line element: $$I = -m \int ds = -m \int \frac{ds}{dt}(q, v) dt = ... 2 You're on the right track. A couple of notes: Those are actually total derivatives. You can think of x(t) and y(t) as functions of t alone you have two equations for two functions. You probably want to isolate them into two equations, each for one function Think about how you would solve this by elimination 2 I assume your car is front wheel drive. The phenomenon is simply Newton's third law in disguise. The car exerts a torque on its forward axle and the wheels exert the same magnitude, opposite sense torque on the car. Normally, the torque is not so big, because as soon as it is exerted on the wheels by the car, the wheels push backward on the road and the ... 2 The use of inertial frames in Lagrangian mechanics is by no means compulsory and everything can be done in any reference frame provided one takes all forces, real and inertial, into account. Actually there are two possibilities in interpreting the question. We work in a non inertial frame R' (instead of an inertial one R) because we are adopting ... 2 You've probably learned about a quantity called "impulse" - try using that to solve the problem. Let J be impulse, defined for constant forces as J = F t where F is the force applied and t is the time for which the force is applied. Since F=ma, we can substitute this to get:$$\begin{aligned} J &= Ft \\ J &= mat \\ J &= m(at) \\ J ...

2

Let's suppose I have some system and I know $M$, the system's total mass, $\vec{r}_{cm}$, the system's center of mass position and $\vec{L}_{cm}$, the systems angular velocity in the frame where the center of mass is the origin. How do I find $\vec{L}'$, the angular momentum with respect to some other origin, say $\vec{r}_{cm} + \Delta \vec{r}$, which is ...

2

You did not carry out your integration quite correctly. We have: $$a(t)= -B_0+B_1t$$ $$v(t) = \int\! a(t)dt=-B_0 t+\frac{1}{2}B_1t^2+C$$ Then plugging in conditions to solve for $C$ we get: $$v(t_s)=0$$ $$0=-B_0t_s+\frac{1}{2}B_1t_s^2+C$$ $$C=B_0t_s-\frac{1}{2}B_1t_s^2$$ Now we can plug in $t=0$ and solve for $v(0)$  ...

1

The simplest setup is for small displacements. Suppose the spring rest lengths are $L_1,L_2,L_3$, the mass has mass $m$, the springs have constants $k_1,k_2,k_3$, the angle is 120 degrees between attachments, and the attachment points are set up so that at rest, the springs are all unstretched. The potential becomes ...

1

The special feature of contact geometry is the contact 1-form $\lambda$, which satisfies $\lambda\wedge d\lambda\ne0$ (let's restrict to 3-dimensions). In our Lagrangian mechanics example, $\lambda = dq-vdt$. You want this to pull-back to zero on the permissible'' curves in phase space -- these curves represent the motions of your system. For a more ...

1

You don't have any time in Your problem, but You can easily calculate it - it is the time it would take to drive 655 meters with 85km/h (I guess that 85km/s in Your question is a mistake), so simply $t=s/v$. If You already calculated the amount of work, simply multiply it by the time calculated, and You'll have the solution.

1

The result is straight forward. As Landau and Lifshitz explain in p.41, when a body disintegrates into two pieces of masses $m_1$ and $m_2$ respectively, their momenta must be equal in magnitude and oppositely directed by the law of conservation of momentum. So, let each body have momentum $p_0$. Then, $(16.1)$ and $(16.2)$ say that the difference in the ...

1

Consider the system in equilibrium, with the rod hanging straight down. Imagine taking a marker and drawing a vertical diameter across the disk. If the rod were fixed to the disk's center so that the disk could not rotate, then would this marker line still be vertical mid swing? What does this tell you about the rotation of the disk around its central axis ...

1

An experimentalist's answer is that the dynamics of all systems are described by second order differential equations, either classical or quantum mechanical. The classical case is called deterministic because all one needs for describing the trajectories in space and time of the particles in the system are solutions of the differential equations . The ...

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