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8

I know that kinetic energy is conserved so $\ \Delta KE=\frac{1}{2}mv^2-\frac{1}{2}I\omega^2=0$ and that $\ I_{end}=\frac{1}{3}Ml^2$ which in this case is $\ \frac{1}{3}*2kg*1m^2=\frac{2}{3}$ You are looking for $\omega (= y)$ from conservation of $KE = (1*3^2/2)$you know that $\frac{1}{2}mv^2-\frac{1}{2}I\omega^2=0\rightarrow 1*3^2 = y^2 ...


8

The correct answer has been given by linus since what causes more damages in an impact is momentum, but the 'soft spot' has been vaguely suggested as a better solution in a comment and in an answer and probably it is opportune to give some concrete and precise information about that issue. The center of percussion is the point on an object where a ...


6

Now, after the hammer is released, the thrower still has her same angular momentum (also 62.8), but the hammer no longer seems to have any. A body does not have angular momentum wrt to a point C only when it is circling around it, you know that planets have elliptical orbits and do have L If a body H has linear momentum p it has also and angular ...


5

I was trying to understand how a photon's angle of reflection is determined, since there really is no such thing as a continuous surface One single atom has no continuous surface, but is made by a nucleus (protons & neutrons) and orbiting electron[s]. You do not need a surface in order to reflect a photon: it can collide with the nucleus or with ...


4

Moving bodies have inertia which means that they will continue to move at a constant velocity unless acted upon by an external force (this is Newton's first law of motion). Similarly, rotating bodies have a moment of inertia, meaning that they will continue to rotate unless acted upon by an external force (torque). Therefore, torque is only required to ...


4

The way you've written them, those are the spin operators in the $\hat{S}_z$ eigenbasis for a spin-1/2 particle. The two $\hat{S}_z$ eigenstates are spin up (written as $|+\rangle$ or $\uparrow$) and spin down ($|-\rangle$ or $\downarrow$), which can be written as $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ in the $\hat{S}_z$ ...


3

First of all you should not think of spin of an electron as a ball spinning on is own axis. At least this comparison should not go beyond your intuitive understanding of an electron spin. The quantity $s=\sqrt{s(s+1)}= \frac{\sqrt3}{2} \hbar$ is an intrinsic property of an electron with spin one-half and has nothing to do with the axis of rotation. I don't ...


3

The diagram is simply a way to remember two variables at the same time, one called the total spin $\hbar\sqrt{3}/2$, the other called the z-component which has two possibilities $\pm\hbar/2$. If you draw a picture of a vector of length $\hbar\sqrt{3}/2$ that is at an angle so that the z-component of the vector is $\pm\hbar/2$ (so it lies on two cones, one ...


3

TL;DR: If you have to choose either "near the handle" or "near the tip", the tip will work better. But there's a point in between these two that works even better; exactly where that point is depends on how you swing the sword, and how its weight is distributed. UPDATED now I am near a computer and can draw diagrams etc. If cutting off the zombie's head ...


3

Consider a particle in a nice Keplerian orbit around a star, taken to be a circle for simplicity. If it is at a radius $r$ and the star has mass $M$, then its specific angular momentum will be $$ l = vr = \sqrt{GMr} $$ and its total specific energy will be $$ e = -\frac{GM}{2r}. $$ If you want to decrease $r$, you are going to have to decrease $l$ as well as ...


3

The propability $P$ that a system in state $|\Psi\rangle$ will eventually transit into state $|\Theta\rangle$ is given by: $$ P = | \langle\Theta | \Psi \rangle |^2 $$ So for example if you have a System in Spin-X-State "up" denoted by $ |\Psi\rangle \hat{=} \frac{1}{\sqrt{2}} (1, 1)^T$, the propability of getting Spin-Z-State "up" (e.g. by measuring along ...


2

In order to get a clear picture consider the outcome of an impact between the head (m = 1Kg) and a blunt rod (plus arm M = 3 Kg, L = 1 m) rotating with angular velocity 1 r/s: If the sword hits with its tip the effective mass is 1 Kg at speed 1 m/s the head would get momentum 1 Kg*m/s If it hits at the hilt (r = .5 m) the effective mass is the same but ...


2

These are mainly conventions. Conventionally, the kets $|+\rangle$ and $|-\rangle$ are taken to be eigenkets of the z-spin operator with, respectively, z-spin of $+\hbar/2$ and -$\hbar/2$. S_x and S_y are chosen such that they obey the canonical commutation relations for angular momenta $$ [S_i,S_j]=i\epsilon_{ijk}S_k $$ E.g., $$ [S_x,S_y]=iS_z $$ and so ...


2

Now, after the hammer is released, the thrower still has her same angular momentum (and has to slow herself down), but the hammer no longer seems to have any. Even though the hammer isn't rotating around the axis, it still has the same angular momentum it had at release with respect to the original axis. So the formula $$L = mvd$$ is correct both for ...


1

Use the hint in your second bullet point. The ball has angular momentum about the pivot point before it strikes the stick.


1

Your conservation of kinetic energy equation should help you solve the for the stick's initial angular velocity. Think of it this way: the tennis ball has initial momentum since it is moving, right? And the stick is not moving, so it has no momentum. At the end of the collision, the tennis ball stops completely, so it has no momentum, but the stick is ...


1

Actually I "believe" that this topic is not really "settled"... The expression $$ \frac12 \left(\sigma_x\otimes\sigma_x+\sigma_y\otimes\sigma_y+\sigma_z\otimes\sigma_z\right) $$ is introduced by P. Dirac in his famous book The Principles of Quantum Mechanics IV ed. chap IX p. 221 where he uses this expression (actually he does not write explicitly the ...


1

Magnetic moment, angular momentum, and charge are related, because the magnetic field is how the electromagnetic interaction carries angular momentum. If there were an intrinsic relationship between magnetic moment and angular momentum, you would expect the neutrino to have a magnetic moment. The current PDG reports an upper limit $\mu_\nu < ...


1

Let us see a similar example: two people on skates going with some velocity towards each other both a bit on left off their common center, and in the moment of the closest approach, they just catch each other by right arms and they start to rotate. In fact they have (as one system) the same angular momentum all the time. When you have a projectile that ...


1

$\newcommand{\ket}[1]{\lvert #1 \rangle}$There is no "reason" you can use the ladder operators. Rather, they are the reason angular momentum is quantized in integer steps. You can define them, there's no inconsistency, so you can use them, and using them leads you to conclude that the angular momentum is raised/lowered in integer steps by them, in the way ...


1

My problem is that I'm not sure if it's right or if it makes sense to keep getting 0 for the uncertainties. Is there some intuitive reason why this makes sense in this context? Yes, it is right. Your $Y_{lm}$ are eigenfunctions of all the operators for which you are calculating the uncertainties. So the uncertainty is necessarily zero! If $$ \hat ...


1

After the mass exits the pipe, the tube will start to rotate from the recoil. At $t=0$, there is zero angular momentum, $L=0$. Let's take it that the pipe rotates about its centre-of-mass, and use that point as the origin from which to calculate the angular momentum. At time $t+dt$, the puff of gas has angular momentum $L_{gas} = dm \times l/2 \times v0$. ...


1

Angular momentum is conserved in this example! As you already stated, the angular momentum of the thrower doesn't change after the hammer is released. Consider the hammer being in rotation around the origin of our coordinate system for $t < 0$: $$ \vec{r}(t) = r_0 \ \ (cos(\omega t), sin(\omega t), 0)^T $$. Its momentum is therefore given by: $$ ...


1

I don't quite understand the construction you are using for $\vec{L_1}$ and $\vec{L_2}$. Normally, $M$ is used to designated the projection of the angular momentum vector along one axis. The idea of his model, consisting here of integer angular momentum, is to consider only the integer values obtained in all the possible vector addition of $\vec{L_1}$ and ...


1

which is by no means integral. What am I missing? Should I round the result to integer? Or does the vector model actually work only qualitatively? They are saying that you should take $L_1$ and $L_2$ to be non-negative integer values. And then $L$ ranges over the non-negative integer values from $L_1+L_2$ to $|L_1-L_2|$. For example, if $L_1=3$ and ...


1

How do the eigenfunctions of the total angular momentum operator analytically look like? I mean the operator is given by $J = L+S$ so the eigenfunctions have to be tensor-product states, right? Can we explicitely say what they are? The eigenfunctions of $J$ are going to be made up of linear combinations of tensor-product states of the ...


1

There are two things going on in the last step. Let's take care of the simpler one first. The third term reduces thanks to a trigonometric identity: $$ -\hbar^2 \left ( 1 + \frac{\cos^2 \theta}{\sin^2 \theta} \right ) \frac{\partial^2}{\partial \varphi^2} = -\hbar^2 \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta} \frac{\partial^2}{\partial \varphi^2} = ...



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