Hot answers tagged

45

Momentum / energy are the conserved Noether charges that correspond, by dint of Noether's Theorem to the invariance of the Lagrangian description of a system with respect to translation. Whenever a physical system's Lagrangian is invariant under a continuous transformation (e.g. shift of spatial / temporal origin, rotation of co-ordinates), there must be a ...


10

As the wiki article you quote states, momentum is defined as the product of the velocity times the mass of an object. Classical mechanics developed theoretically on the lines explained by WetSavanna in the other answer, the conservation of momentum and energy being cornerstones of the theory. Classical mechanics is a very successful theory, and ...


7

This question leads to some subtleties. There are at least two distinct notions of "revolution" that could be meaningful in physics. Namely, "to revolve" can mean: To have angular momentum; To transform by a particular kind of Euclidean isometry (a rotation) (or, to be broader and more technical, a representation of that Euclidean isometry). As far as we ...


7

Without touching on electromagnetism, I'd like to bring up this construction from mechanics (it's in the Feynman lectures). Consider two equal particles approaching each other with equal speed. A----> <----B You can argue from first principles that if they stick together they will not be moving afterwards -- any argument you could make ...


4

Angular momentum, or a measure of rotation, in very large astrophysical or cosmological bodies and energy can become relativistic, and must be treated using General Relativity. In General Relativity (GR) angular momentum is not too different from momentum and energy. Different quantities, the last two are contained in the stress energy momentum tensor and ...


3

It seems relevant to mention the importance of distinguishing between explicit, implicit, and total time-dependence. The Lagrangian $L=L(q,v,t)$ depends implicitly on time via the position $q$ and the velocity $v$. The total time derivative of the Lagrangian $L=L(q,v,t)$ is $$\underbrace{\frac{dL}{dt}}_{\text{total $t$-derivative}}~=~\underbrace{\frac{\...


3

Your reasoning is essentially correct, apart from the last paragraph. To conclude, note that Newton's equation: $$\ddot {\mathbf r}(t) = \mathbf f(\mathbf r (t)),$$ with initial condition $\mathbf x (0)=(x_0,0,0)$, $\dot {\mathbf x} (0)=(0,0,0)$ can be solved by puttin $y(t)=z(t)\equiv 0$, thus reducing to a one dimensional problem: $$\ddot x (t)=f(x(t)),$$ ...


3

What should be the minimum value of $v_0$ in order to hit the monkey while it's in air? Minimum value for $v_0$ is when arrow hits to the monkey just before it (monkey) reaches to the ground. Or, minimum value for $v_0$ is when arrow's range is equal to horizontal distance between hunter and monkey. So you need to find the time of monkey's fall. Or, you ...


2

Often $X$ is a coadjoint orbit of a Lie group. These have a natural symplectic structure; see https://en.wikipedia.org/wiki/Symplectic_reduction


2

Newton's second law, force f is $$f=m\frac{d^2 x}{d t^2}$$ x is position vector of the particle. $$f=-\frac{d v}{dx}$$v is the potential energy. $$m\frac{d^2 x}{d t^2}=-\frac{d v}{dx}$$ Multiply both sides with $\dot x$ $$\frac{m}{2} \frac{d\dot x^2}{dt}=-\frac{dv}{dt}$$ $$ \frac{d}{dt}(\frac{1}{2}m\dot x^2+v)=0$$ i.e., $$\frac{dE}{dt}=0$$Energy is ...


2

The flaw is your assumption that In this [accelerating] frame we don't see any force so the first law of dynamics is respected. In the accelerating reference frame you do see evidence of a force, even though you don't see the effect you are expecting (acceleration of the object). Like an observer standing on the surface of the Earth, acted on by the ...


2

The forces acting on the box will be those by 'gravity', the normal reaction force by the table and the friction force by the table. The force of gravity has two components (passing through the center of inertial mass owing to the equivalence principle) along the surface of the table and normal to it. The friction acts along the surface and the normal ...


2

Mass removal due to wear relates to sliding distance. This gives to wear rate relates to sliding speed. However, wear rate is not only only a function of sliding speed. Surface hardness also plays a role. AL-SI, according to this paper, will be hardened in the beginning session. With the surface hardness increased, the wear rate decreases. When it can no ...


1

Abrasive is intended to increase friction. You want your brake pads to have (and retain) a certain shape, so you use some structure material that resists well to changes in temperature. Not sure what performance material is supposed to be. In order to limit the costs, in places where no structural integrity is threatened, and no friction exists, you can use ...


1

The Lagrangian only depends on the potential energy and the kinetic energy. What the statement you quoted means is that if both the potential and kinetic energies are constant w.r.t. time, then so is the Lagrangian. This makes a lot of sense. Usually, we have: $$\mathscr{L}=K(x,t)-P(x,t)$$ Where $K$ and $P$ are the kinetic and potential energies. But if ...


1

We must all keep in mind that for average atmospheric pressure, and assuming Zhang could pull a hard vacuum with his abdomen (which is probably not feasible, but serves to provide us with a bound), the maximum (negative) pressure he could achieve is only about 14.7 psia. Given 36 tonnes, you can back calculate what the diameter of the bowl would have to be. ...


1

The exact mechanism you describe for how suction cups work is how the rice bowl work. Instead of the bowl being flexible, though, it's his body (skin and muscles) that are providing the change in volume necessary for the suction. So, instead of the suction cup creating the volume change, it's the surface the suction cup (the bowl) is attached to, Mr. Zhang'...


1

From the equations $\dot q = p/m$ and $\dot p=V'(q)$ and the definition $Q=D(q)$ you can derive by differentiation $\dot Q=D'(q)p$ and $\ddot Q=D''(q)pp-D'(q)V'(q)$. The second equation produces an ODE for $Q$ if you can express $p$ in terms of $\dot Q$ up to terms in the null space of $D''(q)$ (i.e., translation and rotation degrees of freedom). This should ...



Only top voted, non community-wiki answers of a minimum length are eligible