Hot answers tagged

4

We only need worry about angular momentum when calculating other angular behaviors. In your case, you want to find the linear speed (velocity but without worrying about direction) of each ball. The speed of a point on a circle is the angular speed in radians/second multiplied by the radius of the circle, $2\pi r$ (to get to circumferences traveled per ...


3

Refs. 1 and 2 only discuss a generic/bulk notion of a constant of motion in phase space; not any refinement thereof restricted to a subsurface of phase space. Let us here carefully explain the generic/bulk notion of a constant of motion used in Refs. 1 and 2. Definition 1: An on-shell constant of motion $F(q,p,t)$ satisfies $$\tag{1} ...


3

When you say that $L_x$ and $L_y$ vanish for a point confined to move in the plane $z=0$, you mean that the the solution $\vec{x}=\vec{x}(t)$, $\vec{p}=\vec{p}(t)$ describes a curve in the given plane with tangent vector parallel to that plane. So that, exactly along that curve, $$L_x(\vec{x}(t),\vec{p}(t))= L_y(\vec{x}(t),\vec{p}(t))=0\quad \forall t \in ...


3

One can use the Doppler effect, which will shift spectral lines to the red at the side the rotates away from us and towards the blue on the side that rotates towards us. This is being used by astronomers who measure "rotational broadening" on stars which can not be resolved in telescopes. In that case it's all about measuring the rate of rotation, of course, ...


3

The picture of a black hole being like a huge vacuum is pretty misleading, not to say wrong. The gravitational field around the black hole is exactly the same as around any other object of the same mass. Like any other object, it's perfectly possible to orbit a black hole, just like we orbit the Sun and don't fall in. In fact, if you were to magically ...


2

You could try measuring the effects of the Lense-Thirring effect. It is an example of frame dragging. Essentially, an object that is orbiting near a massive object that is also rotating will have its axis undergo a change in its orientation. There are two problems here: The rotating object must be large. The rotating object must not be rotating slowly. ...


2

We generalize the result from the second answer you linked: $$\vec{L}=\sum_i \left(\vec{r}'+\vec{{r}_i}' \right)\times \vec{p_i}=\vec{r}' \times \sum_i \vec{p_i}+\sum_i \vec{{r}_i}'\times\vec{p_i}=\vec{r}' \times \sum_i \vec{p_i}+\vec L' $$ Now $\vec{r}'$ is some arbitrary vector, not just the vector to the center of mass. Thus when the system is moving ...


2

The torque exerted by $\vec B$ is perpendicular to it, so the $z$ component of angular momentum is conserved.


1

This is not my answer, it's one of the answers you can find here Is there a reason why the spin of particles is integer or half integer instead of even and odd? I just wrote here (and re-posted) the work of @Siva which I found a very good answer. However, follow the link to read more interesting useful answers The "spin" tells us how the wavefunction ...


1

yes i think your assumption is correct based on the fact that in equilibrium the discs shouldn't slip on each other. but then you need to change your equation for angular momentum. use parallel axis theorem to write down the moment of inertia for the stationary disc to get the answer in the equation. since you have chosen your axis to be passing through the ...


1

When you drop a stationary disc onto a rotating one there must be a time when there is relative motion between the discs as you cannot have an infinite acceleration. If there is no friction then nothing much happens and the spinning disc carries on spinning and the other disc just sits still on top of it. To get an interaction between the discs you need ...


1

The intuitive explanation is that an eigenstate of $\hat L_x$ with eigenvalue $m_x=0$ is invariant by rotation around the $x$ axis (by definition), but not by rotations around any other axis. Thus it cannot be an eigenstate of $\hat L_z$ with eigenvalue $m_z=0$. See for example this picture of the eigenfunctions of $\hat L{}^2$ and $\hat L_x$ :


1

I mean, if you believe what you've just written, $$\theta(t_2)-\theta(t_1)=\theta(t'_2)-\theta(t'_1)$$ Then I think you have already proven it, with a little more formal calculus. First rewrite that expression above as $\Delta \theta(t)=\Delta \theta(t')$, with the prime indicating the other interval. If $$t_2-t_1=t'_2-t'_1,$$ Then just set $\Delta ...


1

Since you know about $SU(2)$ characters, this is a doable exercise. Let me remind you that the character of spin-$j$ is $$\chi_j(t) = \sum_{m=-j}^j e^{2imt} = \frac{\sin (2j+1)t}{\sin t}.$$ The crucial property is that the character of a tensor product is the product of characters, i.e. $\chi_{j \otimes j'}(t) = \chi_j(t) \chi_{j'}(t).$ In your case, ...


1

Define $D=\{(x,y)\in\mathbb R^2:\|(x,y)\|\leq R\}$ as the disk of interest. There are two spaces of interest here: the space of square-integrable functions on $D$, $L_2(D)$, and the space of such functions with Dirichlet boundary conditions, $\mathcal H=\{\psi\in L_2(D):\psi(p)=0\:\forall p\in \partial D\}$. You're interested in the hamiltonian ...


1

In order for the spinning disc to set the stationary one in motion, friction forces need to act between the contacting surfaces: That requires a Normal force $F_N$ to act between them. Using the kinetic friction coefficient $\mu_k$ we can then state: $$F_F=\mu_k F_N$$ The friction force on the $m,r$ disc causes angular acceleration $\alpha=\frac{d ...


1

Yes you have to take the frictional force F into account so this exerts a negative impulse moment RFdt on the right hand disk with radius R as well as rFdt on the left hand disk with radius r. So the delta of the rotational moments have the ratio r/R. Assuming the left hand disk spins with rim velocity v0 before impact and after impact they have equal rim ...


1

"Say we have two spinning spheres, each spinning about an axis going through their own centre of mass. I have highlighted a very important statement in your question. If you have a body rotating about its centre of mass then if the centre of mass does not move the angular momentum of the body will be the same about any point. So the angular momentum of ...


1

The impulse is the change in total momentum of the body of mass. The momentum $P$ of a rigid body is the product of the velocity of the centre of mass $v$ and total mass $M$. A free rigid body's centre of mass translates at constant velocity, while the body itself performs a rotation about the centre of mass. Thus, after the impulse $J$ has been applied, ...


1

There is a pattern. If you fix the eigenbasis of $S_z$ you have for any up $+$ or down $-$ eigenstate in $\vec{u}$ direction: $$ \vec{u}=\sin\left(\theta\right) \cos \left(\phi\right)\vec{x}+\sin\left(\theta\right) \sin \left(\phi\right)\vec{y}+\cos\left(\theta\right)\vec{z} $$ then $$ S_u=\sin\left(\theta\right) \cos ...


1

Pouring anything into the cup adds mass but doesn't add angular momentum. Conservation (plus friction between fluid and cinnamon) of angular momentum requires the new total mass to rotate slower than the original contents.


1

We know that $\psi_{l,m}$ satisfies, for each $l$ and $m$, the equations $$L^2\psi_{l,m}(r,\theta,\phi)=l(l+1)\hbar^2\psi_{l,m}(r,\theta,\phi),$$ $$L_z\psi_{l,m}(r,\theta,\phi)=m\hbar \psi_{l,m}(r,\theta,\phi).$$ But we also know that, by definition, $Y_{l, m}(\theta, \phi)$ satisfy the same equations. Then, unless the eigenvalue equations for $Y_{l,m}$ ...


1

If a ball of say radius $R$ rolls without slipping it has both linear ($p$) and rotational momentum ($L$): $$p=mv$$ $$L=I\omega$$ Where $m$ is mass of ball, $I$ is inertial moment of ball, $v$ is translational (linear) speed and $\omega$ is angular speed. For rolling without slipping the following condition also holds: $$v=\omega R$$ The ball will have ...



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