# Tag Info

## Hot answers tagged classical-mechanics

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 ...

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 ...

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 ...

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 $\mathbb{R}\times \mathbb{R}^3$ ...

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.

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

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 ... 1 I) Many of OP's questions on how the Lagrangian formalism works is already addressed in e.g. this Phys.SE post and links therein. For instance the question about the total time derivative in the EL equations is discussed in my answer. II) In this answer, we would like to explain mathematically the various definitions in the Lagrangian formalism (of ... 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}}$. The ... 1 Look at sparknotes.com/physics/specialrelativity/dynamics/…, you can see$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 ...

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