# Tag Info

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A pure moment cannot be created by one force. In real life, producing pure torques is almost impossible. There will always be a non-zero net force applied also. In your examples the torque about the center of mass is $\tau_C = 2 r F$ in both cases, but on the second case you also have a net force applied $2 F$ that changes linear momentum also. So your ...

1

The error lies in the theoretical confusion between forces/torques and energy. The kinetic energy is linked to the motion generated by forces and torques which are the causes of the motion itself. Understanding the energy value in the two situations results impossible without knowing the time course of the applied force, being the energy, and so the work, ...

0

www.physicspages.com/2013/04/11/magnetic-dipole-moment-of-spinning-spherical-shell/ My search gives $\mu = \frac{e\omega R^2}{3}$ This gives $g = 5/3 = 1.667$ Did not you provided link given below? https://en.wikipedia.org/wiki/Electron_magnetic_moment#The_classical_theory_of_the_g-factor Which explains that non-uniform charge distribution can explain ...

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Kinetic energy is $$K = \frac{1}{2} \vec{\omega} \cdot [I] \vec{\omega}$$ when the 3×3 mass moment of inertia matrix $[I]$ is expressed in world coordinates. Remember $$[I] = [R] [I_{body}] [R]^\top$$ is how body inertias is transformed into world inertias. You seem to apply a scalar mass moment of inertia to a vector rotation. If you are careful with the ...

0

The multiplicity $2S+1$ actually tells you how many degenerate spin states there are, each labelled with the total spin projection quantum number $M_S$ (this is from the total spin projection operator $\hat{S_z}$(conventionally taken to be in the z-direction) whose eigenvalues are $\hbar M_S$). The possible values of $M_s$ are $-S\le M_S\le S$ in integer ...

1

In general, the change in angular momentum resulting from a change in moment of inertia depends on how the change is implemented, and to some extent your perspective. In physics, you can think of global conservation laws as constraints that feed into your interpretation of a system. Consider the simple problem of determining the change in linear momentum ...

2

The principle of conservation of angular momentum says that angular momentum remains conserved unless an external torque acts on it. The net torque on a body is defined as: $$\vec{\tau\,}=\dfrac{\mathrm d\vec{L\,}}{\mathrm dt}$$ We can clearly see from this definition that since external torque on the body is zero, the angular momentum is going to remain ...

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Remember that the variation of the angular momentum equals the external torque. If there are no external torque (as in your case), the angular momentum is conserved.

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The notation $(\vec S_1\otimes 1) \cdot (1\otimes \vec S_2)$ is interpreted as a contraction over spatial indices. Hence, $\vec S_1\cdot \vec S_2=\sum_{ij} \delta_{ij}S_{1i} \otimes S_{2j}$. By contrast, $(S_{1x}+S_{1y}+S_{1z})\otimes(S_{2x}+S_{2y}+S_{2z})=(\vec S_1\cdot [1,1,1])\otimes (\vec S_2\cdot [1,1,1])$.

0

The loss of energy (efficiency<1) manifests itself in two ways Friction. When doing free body diagrams add frictional forces at the sliding contacts and find which coefficient of friction $\mu$ gives you the efficiency values you expect. Structural damping (hysteresis). This is more difficult to put in the equations of motion, because they assume ...

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You have my sympathy; precession is little understood. Try it this way! i) remove the front wheel from a bicycle and mark a spot on the tyre. ii) hold it by the ends of the axle between your hands with the mark at "top dead centre". iii) set the wheel spinning as though the machne was progressing forward (very slowly) Note that the spot travels ...

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Your feeling looks very misguided. Whatever you do, stay away from SU(3) for rotations. The rotation group and its Lie algebra are always linked to SO(3) ~ SU(2), to avoid formal forays into double covers and half angles. Read up on the spin matrices for any representation of the very same group (any spin). There are, in fact, simple systematic ...

1

MRI signal is always complex and it is related with signal demodulation. The detected signal is multiplied by a sinusoid or cosinusoid with frequency equals to $\omega_0 +\delta \omega$, respectively leading to the real and imaginary channels. You can find the complete algebra at $Haacke,\ Magnetic\ resonance\ imaging$ chapter 7.3.3 Phase is really useful ...

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First of all, the transmission of force between any two pairs of mating gears happens due to surface contact between the two and hence through friction. Next, any loss is a dissipation in energy. Though I am not that well-versed in vibration analysis, I know that an periodic dissipation of energy is accounted in the force-equation using a damping ...

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The space ship should undergo both linear motion and angular (rotational motion). Linear motion will take place in the direction in which the force is applied, and rotation about its centre of gravity. So the centre of gravity will act as an imaginary pivot.

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a Stern Garlach apparatus does not rotate the state of the particle, what it does is to split a beam or, if you have a single particle, it chooses" a state in the desired direction. What you might be looking for is how to write the eigenstate in terms of Z basis. For a 1/2 spin it is going to be: $$|\pm \rangle_y = \frac{1}{\sqrt{2}}\left( |+\rangle_z ... 3 The way to do this is using the Wigner-Eckart theorem. The way it is applied to your problem is as follows:$$ \left\langle nlm |\vec{r}| n'l'm'\right\rangle = \left\langle nl ||\vec{r}|| n'l'\right\rangle \left\langle l' m' 1 q | l m\right\rangle $$where the second factor is a Clebsch-Gordan coefficient and q=-1,0,1 indicates the type of transition. For ... 1 I) For the record, here is the operator calculation that OP wants to avoid. The benefit of the calculation is that the operators are not sandwich with any bra/ket representation, and hence we do not have to worry about whether the bra/ket representation is faithful. Let us put \hbar=1 for simplicity. The starting point is the CCR$$ [x^i, ...

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So the definition of angular momentum should be this: "Angular momentum is the product of the angular velocity of the body or system and its moment of inertia with respect to the rotation axis, and that is directed along the rotation axis". That's not a useful definition at all, because (i) it does not specify what this "moment of inertia" thing is, and ...

2

I am certain that mathematical analysis of tidal locking has been done many times but I have failed to find such an analysis where the mechanism for the transfer of angular momentum to spin angular momentum is included which is the question which has been asked. Perhaps someone is able to produce a reference or an analysis? Having experienced on a number ...

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Anna V gave the correct explanation. Only when weak bosons are created on mass shell, e.g. at collisions @ Ecm = M, can you apply total angular momentum conservation at a single vertex (production and decay). On the other hand, even for off mass shell bosons chirality (read helicity for ultra relativistic particles) imposses costraints at each vertex.

1

The concept of angular momentum only makes sense when we specify a rotation axis. So we will pick the axis passing through the initial center of rotation. When the string is cut, the point mass has a linear momentum $p = mv = mr\omega$. One can define the angular momentum of a particle about an axis as $\vec{L} = \vec{d} \times \vec{p}$ where $\vec{d}$ ...

4

I think that what your teacher has told you is that the angular momentum of a body can be split into two components: The spin angular momentum which is an intrinsic property of the body and is independent of the point about which you wish to find the angular momentum. $L_{\text{spin}} = I_{\text{cm}} \omega = \frac v r$ where $I_{\text{cm}}$ is the ...

0

There is an exact solution to the problem. You can consider each point on the sphere by specifying two parameters ($r$ and $\theta$). Now since you know each points velocity vector you can calculate its contribution to the angular momentum by using $dL$ = $dm$($r$ x $v$). Now integrate over $\theta$(0 to $2\pi$) and $r$(0 to R). After doing all this, you ...

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In short, you must consider the total elements of the system for conservation of momentum. In this case, nearly all of the momentum is exchanged between the electron and a photon that is absorbed or radiated away (the light). Momentum is conserved, and is largely balanced by this electron-photon interaction, although smaller amounts may be exchanged with ...

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Angular momentum is conserved only if there's no external forces, in this case the electron gains energy by light or by heat wich is kinetic energy. They are both external forces so the conservation of angular moment does not apply.

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The answer is two arcs. One arc with a constant gee loading in one direction and then flipping to the opposite direction. This is called the bang-bang method, and it is no very smooth, but the gee forces never exceed the specified maximum. Given a path $y(x)$ the instantaneous radius of curvature at each x is $$\rho = \frac{ \left(1+ \left(\frac{{\rm ... 0 The cross product of two vectors is the area of a parallelogram defined by those vectors. If the angle is 90 degrees then the cross product is simply the magnitude of vector 1 multiplied by the magnitude of vector 2 (i.e are of a rectangle). another way of writing the cross product is |v||u|sin(theta). If theta is 0 (both pointing in same direction) or 180 ... 0 Bohr's model is the best you can do as a classical analogy for discrete quantum bound states. There is no classical analogy that explains how the \ell=0 state manages not to collide with the nucleus. And there's a good reason that there is no classical explanation for this: The fact that the electron in the ground state of hydrogen does not "fall to the ... 1 It should be easy to see that if it "disintegrates" into dust particles, it is no different than if you have a rotating object, held by a string, and the string is cut. We know that the angular velocity (momentum) \omega will be converted into a tangential velocity v and the particles will have a momentum mv, in addition to the momentum generated by ... 0 Upon a rotation, that matrix element is equivalent to$$ \langle +,z|S_x|+,z\rangle $$As S_x=a S_-+b S_+ for some a,b\in\mathbb C, the matrix element is trivially zero, because \langle +|-\rangle=0. Or put it another way: you can always relabel the axes so that x\leftrightarrow z, in which case the matrix element would look like \langle ... 1 This is indeed confusing. The confusion comes from this very peculiar hypothesis: What if the person doesn’t apply a tangential friction force at his feet? It implies there is a radial contact force at the person's feet (I prefer "contact" to "friction", which refers to movement). And, indeed, for the person to move radially inwards, or even to stay ... 1 The vector \mathbf r does change, even though its magnitude is nearly constant. Most importantly, the component of \mathbf r which is perpendicular to the rotation axis is decreasing in this example. This fact leads to the explanation for which you are searching. Another way of seeing this would be to use this definition of angular momentum, which ... 1 You have to approach problems systematically, and not intuitively. Like I stated in a previous (accepted) answer, resolve everything on the center of mass, and only in the end transfer the quantities to a different point (like P) to get the results you want. I start with the kinematics. Use \ell_1 and \ell_2 for the horizontal distances and h for the ... 2 If you draw similar triangles, then you'll find that r_A/r_B = v_y/v_x, and so the product r_A v_x is equal to r_B v_y. Try drawing a line from the tip of your lower \vec{v} vector to the tip of your lower v_y component to see this. 1 For linear momentum to be a constant, its value should not change. Being linear momentum a vector quantity, it's value is completely described by specifying both magnitude and direction. In a circular motion, even though the speed remains a constant, the direction of velocity (which is tangential to the point on a circle) is changing throughout the motion as ... 0 Does this mean that without any external force the kinetic energy of any thing can be increased? No, of course not. Just because there is no torque that doesn't mean that there isn't a force. In this particular case, you have an inwards radial force that is performing work by countering and exceeding the centrifugal force. This work is what causes the ... 2 As you mentioned, \vec {L_o}_⊥ is proportional to \omega. So the middle term means \vec {L_o}_⊥ varies by varying angular velocity. You can also derive this. formula like this:$$\frac{d \vec {L_o}_⊥ }{dt}= \frac {d\vec{\omega}}{dt}A \hat{L_o}_⊥+ \vec{\omega}A \frac{d\hat{L_o}_⊥}{dt}= \frac{1}{\omega} \frac{d \omega }{dt} \vec{L_o}_⊥ + \vec{ \omega } ...

2

Yes, there are more unknowns than equations. You do not have sufficient information to solve for the requested quantities. Someone might be playing a prank on you! In reality, each ball and each paddle would have a specific finite stiffness, and one could use this information along with some clever math to determine the final velocities of all the bodies. ...

2

Let the body rotate about the $z$-axis, then by the definition of angular momentum $$\vec{L}=\vec{\omega} I_z.$$ where $\omega$ is the angular velocity about the $z$-axis. So we could take the parallel axis theorem and multiply it by $\omega$: $$\vec{\omega}I_{z}=\vec{\omega}I_{cm}+\vec{\omega}ma^2$$ Now ponder the terms in it. If I understand the ...

1

You forgot to include angular momentum of center of mass. Thus add $MR^2\omega$ to your answer By the way angular momentum is not $I\omega$, it is $\vec{l_{com}}+I_{com}\vec{\omega}$ It is $I_{contact}\omega$ only for cases when point of contact is the point of instantaneous or fixed axis of rotation.

0

In my view the motion of the person, seen in lab frame, would be linear motion, because, at the beginning, the person has a tangential velocity $ωr$ and radial velocity $v$, and he will keep these two forever. But then does it makes sense to talk about conservation of angular momentum? I mean it will surely be conserved in the lab frame but the ...

0

After reading your question again I gave a try proving why and how angular momentum is conserved. In the example given by the question one needs to understand how the angular momentum of an object moving in a rotating, non-inertial frame is conserved. I will present in brief, my effort of proving how and why the angular momentum is an integral of motion. The ...

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The important factor here is the precise definition of the constraint at O. From your description of what the textbook describes occurring, I take it that the rod is supported at O by a constraint which allows rotation, but only about some axis which is part of the definition of the constraint (a hinge, not a ball joint). Such a constraint can exert ...

2

You don't have to, but it makes the equations easier to deal with because you don't have to account for the moment of acceleration terms. See the 2nd part this this answer about deriving Newton's laws on an abitrary point not the center of mass. So finally the equations of motion of a rigid body, as described by a frame A not on the center of gravity C ...

0

The axis of rotation changing even when there is no applied torque is an effect which is possible in 3-dimensions but not in 2-dimensions. The angular momentum about an axis $\vec L$ is given by $\vec L = I \vec \omega$ where $I$ is the moment of inertia about the axis and $\vec \omega$ is the angular velocity about the axis. If you are observing an ...

1

A rigid body moving with no constraints, in particular rotating, will rotate necessarily about a principal axis of inertia. I thought that the reason of this is that otherwise, the angular momentum $\vec L$ would not be parallel to the angular velocity $\vec\omega$, hence it would follow a precession motion and that would imply the presence of a ...

1

The answer is the parallel axis theorem. It states that the inertia tensor $I_{ij}$ will transform under a translation with vector $\vec{a}$ as $$I = I^{(\text{cm})} + m \begin{pmatrix} a_2^2 + a_3^2 & - a_1a_2 & -a_1a_3 \\ -a_1a_2 & a_1^2+a_3^2 & -a_2a_3 \\ -a_1a_3 & -a_2a_3 & a_1^2+a_2^2\end{pmatrix}\\$$ where $I$ is the new ...

1

As Praan confirmed, you are fine. Your result is the n=4 case of the general expression for composing spin 1/2 doublets, (so here denoted by their dimensionality, 2), $${\mathbf 2}^{\otimes n} = \bigoplus_{k=0}^{\lfloor n/2 \rfloor}~ \Bigl( {n+1-2k \over n+1} {n+1 \choose k}\Bigr)~~({\mathbf n}+{\mathbf 1}-{\mathbf 2}{\mathbf k})~,$$ where $\lfloor n/2 ... 1 The error is just to consider an average speed$h\omega$. When the particle is at height$z$, its horizontal (relative to the Earth) speed is$v=2z\omega$. The time of of flight is $$t=\sqrt{\frac{2z}{g}}.$$ Differentiating this expression we get the time taken by the particle to move a distance$dz\$, $$dt=\frac{dz}{\sqrt{2gz}}.$$ The horizontal distance ...

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