For uniform charge and mass distribution for a rigid body rotating with a uniform angular velocity about its centre of mass, I want to know if it is just a mathematical artifact of integration which cancels out the factors of moment of inertia (angular momentum) and the magnetic dipole moment or if there is something much deeper. I realize the spin magnetic moment $g$-factor is 2 quantum mechanically, which is another reason why i want to know if there is some kind of emergent property which nicely cancels everything and brings the final result as $q/2m$ in classical physics
57315
How do I prove the relation between the vectors of magnetic moment $\vec\mu$ and angular momentum $\vec L$, $$\vec\mu=\gamma\vec L$$  ?  Many text books and lecture notes about the principles of magnetism show the relation of $\mu$ and $L$ as scalars only and then just state that the relation holds also for the vectors. An example:   $$\mu=I\cdot A = \frac{q}{t}\pi r^2 = \frac{qv}{2\pi rm}m\pi r^2$$  ($I=q/t$: current, $A$: area of a loop, $q$ charge, $t=2\pi r/v$: time of 1 rotation, $v$: velocity of particle, $m$: mass )  The angular momentum is $\vec L = \vec r\times\vec p$ or $L=mrv$ and therefore $$\Rightarrow \mu = {\frac{q}{2m}} L = \gamma L.$$  Why is this also true for the vectors? Is there a general explication by classical physics without the need of quantum theory?
24170
This may seem like a slightly trite question, but it is one that has long intrigued me.  Since I formally learned classical (Newtonian) mechanics, it has often struck me that angular momentum (and generally rotational dynamics) can be fully derived from normal (linear) momentum and dynamics. Simply by considering circular motion of a point mass and introducing new quantities, it seems one can describe and explain angular momentum fully without any new postulates. In this sense, I am lead to believe only ordinary momentum and dynamics are fundamental to mechanics, with rotational stuff effectively being a corollary.  Then at a later point I learned quantum mechanics. Alright, so orbital angular momentum does not really disturb my picture of the origin/fundamentality, but when we consider the concept of spin, this introduces a problem in this proposed (philosophical) understanding. Spin is apparently intrinsic angular momentum; that is, it applies to a point particle. Something can possess angular momentum that is not actually moving/rotating - a concept that does not exist in classical mechanics! Does this imply that angular momentum is in fact a fundamental quantity, intrinsic to the universe in some sense?  It somewhat bothers me that that fundamental particles such as electrons and quarks can possess their own angular momentum (spin), when otherwise angular momentum/rotational dynamics would fall out quite naturally from normal (linear) mechanics. There are of course some fringe theories that propose that even these so-called fundamental particles are composite, but at the moment physicists widely accept the concept of intrinsic angular momentum. In any case, can this dilemma be resolved, or do we simply have to extend our framework of fundamental quantities?
48693
I know that when a body is is equilibrium   There is zero resultant force in any direction, i.e. the sum of all the components of all the forces in any direction is zero The sum of the moments about any point is zero (this point does not need to be on the body)   What if the body is not in equilibrium?  Let me clarify: What if there is a resultant moment of $5\text{ Nm}$ clockwise about a point on a body. If I calculated the moment about another random point in or outside the body, would I still get $5\text{ Nm}$ clockwise as my value?
3136
I am reading An Introduction to Mechanics by Kleppner and Kolenkow (2014). On page 241 is the definition of the :      Here is the formal definition of the angular momentum $\vec{L}$ of a particle that has momentum $\vec{p}$ and is at position $\vec{r}$ with respect to a given coordinate system: $$\vec{L}=\vec{r} \times \vec{p}$$   In the book there is no explanation why this formula should be true. From this equation the formulas for torque and moment of inertia are derived.   My question is: Why is the formula above correct? Why isn't the formula for angular momentum something completely different, like $\vec{L}=\sqrt{(\vec{r} \times \vec{p})2\pi M}$? Is the formula for angular momentum just a mere definition and if not, how to derive it? How did people come across that particular formula?
27824
I understand that angular momentum is a vector, etc..  But, what really happens when some object, say a ball for example, is set to rotate along two axes? What would the resulting motion look like?
67423
Suppose we have a particle of mass $m$ confined to the surface of a sphere of radius $R$. The classical Lagrangian of the system is   $$L = \frac{1}{2}mR^2 \dot{\theta}^2 + \frac{1}{2}m R^2 \sin^2 \theta \dot{\phi}^2 $$   The canonical momenta are  $$P_\theta = \frac{\partial L }{\partial \dot{\theta }} = m R^2 \dot{\theta }$$ and $$P_\phi = \frac{\partial L }{\partial \dot{\phi }} = m R^2 \sin^2 \theta \dot{\phi }$$  The Hamiltonian is   $$H = \frac{P_\theta^2}{2 m R^2} + \frac{P_\phi^2}{2 m R^2 \sin^2\theta }$$  Now start to quantize the system. We replace $P_\theta $ and $P_\phi $ as $-i\hbar \frac{\partial}{\partial \theta}$ and $-i\hbar \frac{\partial}{\partial \phi} $, respectiely, to obtain   $$H = -\frac{ \hbar^2 \partial^2}{2 m R^2 \partial \theta^2} - \frac{\hbar^2 \partial^2 }{2 m R^2 \sin^2\theta \partial \phi^2 } $$   This is apparently wrong, it should be the total angular momentum!   So what is the right procedure to quantize a system, especially a system in curvilinear coordinates?
111715
A uniform rod of length L and mass M is free to rotate on a frictionless pin passing through one end. The rod is released from rest in the horizontal position.    What is its angular speed when it reaches its lowest position?   I don't understand why we use the center of mass, since this object is constrained by the pivot, wouldn't it be correct to use L instead of 1/2 L?  I got w = sqrt(6g/L) instead of w = sqrt(3g/L)
69255
While studying rotational mechanics, I came across a section where it mentioned that angular momentum may not necessarily be parallel to angular velocity. My thoughts were as follows:  Angular momentum ($L$) has the relation $L=I\omega$ where $\omega$ is angular velocity and $I$ is the moment of inertia, so following this relation, it seems they should be in the same direction. Why are they not?
37082
I haven't touched Physics and Math (especially continuous Math) for a long time, so please bear with me.  In essence, I'm going over a few Physics lectures, one which tries to calculate the Force exerted by uniform magnetic field on a semi circular current carrying wire.  The mathematics that puzzles me is this, that:  $$ \sin(d \phi) \thickapprox d\phi $$  where $d\phi$ is very small. .
14741
Consider a cylinder that rotates about a vertical fixed axis with angular velocity $\vec{\Omega}$ while rotating about  a vertical axis passing through its center of mass with angular velocity $\vec{\omega}$.    Now the two angular velocities of rotation $\omega$ and $\Omega$ are not the same I suppose and there is no link between the two angular velocities, is that right?  Firstly what is the angular velocity of the rigid body? I suppose that it is  $$\vec{\omega^{*}}=\vec{\omega}+\vec{\Omega}$$  But why?  And then what is the moment of inertia with respect to the axis $z$ of the rigid body? If I use parallel axis theorem I get  $$I_{z}=I_{cm}+mR^2$$  Where $R$ is the distance from $z$. But that doesn't seem right. The cylinder is not rotating about its center of mass at the same angular velocity of rotation about $z$. So using $(1)$ give some incorrect results. For istance kinetic energy  $$K=\frac{1}{2} I_{z} \vec{\Omega}^2= \frac{1}{2}(I_{cm}+mR^2)\vec{\Omega}^2$$  But that's the kinetic energy of a body rotating about $z$ and center of mass with the same angular velocity $\vec{\Omega}$, and this is not the case.  I'm confused, can anyone help me with give suggestion about angular velocity and moment of inertia in problems where the rigid body is rotating at the same time about two different axes?
71887
I was reading up on the  (from Sean Carroll's book) and something that he said confused me.   To start with, the Kerr metric is pretty messy, but importantly, it contains two constants - $M$ and $a$. $M$ is identified as some mass, and $a$ is identified as angular momentum per unit mass. He says that this metric reduces to flat space in the limit $M \rightarrow 0$, and is given by $$ds^2 = -dt^2 + \frac{r^2 + a^2 \cos^2\theta}{r^2 + a^2}dr^2 + \left(r^2 + a^2 \cos^2\theta \right)d\theta^2 + \left(r^2 + a^2 \right)\sin^2\theta d\phi^2 $$  and $r$, $\theta$ and $\phi$ are regular spherical polar co-ordinates.   But I don't understand why this space is obviously flat. The Schwarzschild metric also contains terms involving $dt^2$, $dr^2$, $d\theta^2$ and $d\phi^2$ but is curved. I always thought that a metric with off diagonal elements implied a curved space, but clearly I was very wrong.  Question: How do you tell if a metric is curved or not, from it's components?
71268
My understanding of the Stern–Gerlach experiment is that neutral (0 total charge) particles are sent through a non-homogeneous magnetic field, with the expectation that the field will push that particle's path higher or lower on a detector because of the collective spin of that particle. While the detector can detect particles in a two-dimensional surface, the results of the experiment are that particles appear in only two localized areas directly above the path of the particle stream, or directly below the path of the particle stream - spin up or spin down. The conclusions from these measurements are that the particle, when measured, will always have about the same magnitude.  Why does this not follow from classical mechanical theory related to magnetism? If you shot a magnet through a similar apparatus, I would expect the magnet to be rotated to align with the magnetic field in some way which, at high enough field strengths relative to the mass of the magnet, would cause us to measure basically the same magnitude as if the magnet entered the apparatus pre-aligned with the field.  How is my explanation incorrect?
123891
This is the excerpt from my book, Arthur Beiser's Concepts of Modern Physics:      An atom with a certain value of $\displaystyle{m_l}$ will assume the corresponding orientation of its angular momentum $\mathbf{L}$ relative to an external magnetic field if it finds itself in such a field. However, we note that $\mathbf{L}$ can never be aligned exactly parallel or anti-parallel to $\mathbf{B}$ because $L_z$ is always smaller than $\sqrt{l(l + 1)\hbar}$ of the total angular momentum.   Why can't $\mathbf{L}$ be parallel to the applied magnetic field? I am not getting the reason showed by the author. Can anyone help me conceive the reason stated above?
70844
I found after searching that this question has been asked before. But all the answers were not convincing. Suppose I have a body which is free, not constrained always rotate about its center of mass (COM). Why is that so? A convincing answer that I found was that in most cases the moment of inertial about the center of mass is the least and that's why the body rotates about the center of mass. But I ask it again with hope of the question not getting closed and getting a better succinct answer. I was thinking that motion about the COM is the most stable one and the rotation about other points degenerates. I don't think it's right. Is it? —————————————————————————————————— Note-: 1). This question has been wrongly closed. The other questions linked don’t answer my question at all. It asks me to ask a new question if my question is still not resolved. I did make it clear that I am not satisfied with the answers in the linked questions. 2). The answer to this question is that a free body never rotates about its center of mass ( the instantaneous axis of rotation never passes through the center of mass). In fact we choose a point about which we want to decompose the motion into rotation and translation and we could very well have chosen any point other than the center of mass and analysed rotation about it. Moreover the instantaneous axis of rotation for a free body never passes through the center of mass. I would urge the moderators to give me the right to add my answer to this question. This is the correct answer, the one which satisfied me the most and it is nowhere in the linked answers. So kindly give me the right to open this question and let me add my answer to it.
83852
When I am on earth, the weight of my body is countered by the reaction of the ground. So, there is no net force acting on me.  But I am spinning with earth. But if there is no centripetal force then why am I spinning? And the equal air pressure on both side of my body won't be enough for me to stay in the same angular velocity as the earth.  Is it just conservation of angular momentum?
7063
How do I calculate the moment of inertia of a uniform solid cube about an axis passing through its center of mass?  I also wanted to know if the moment of inertia of a body is independent of its shape. Also, recently I read somewhere that the moment of inertia of a  uniform solid cube is minimum about an axis passing through its COM because the mass is more concentrated at its center. Does the statement make any sense?
47850
We know that angular momentum is defined as the cross product of position and linear momentum. By taking the time derivative, we can deduce that the time rate of change of the angular momentum equals the net torque.  Then why is the conservation of angular momentum considered a law, when we can easily show it is conserved when no net torque is applied, by using Newton's laws?  It truly confuses me.