# Tag Info

1

Yes, you can use the eigenstates of $J_y$ in a very simple way: For $j=1$ as you have, let $|1, \mu\rangle$ be the eigenstates of $J_y$ ($\hbar = 1$), $$J_y|1, \mu\rangle = \mu |1, \mu\rangle\;\;, \mu = -1, 0 , 1\\ |1, \mu\rangle = a_\mu |1,-1\rangle + b_\mu |1,0\rangle + c_\mu |1, 1\rangle$$ Then the states $|1, \mu\rangle$ are also eigenstates of ...

3

The tangential acceleration $a_t$ and the angular acceleration $\dot{\omega}$ are basically the same thing. They are related by: $$a_t = r\dot{\omega}$$ So we don't include both of them because that would be counting the same thing twice.

0

The beam will be accelerating downwards after one guy lets go, and it will be rotating. I would ask myself - how fast is it accelerating? How fast is it rotating? There will be a torque $\Gamma$ due to the remaining force $F$ of the one man $\Gamma=F\ell/2$ resulting in angular acceleration $\dot \omega = \Gamma/I$; and a vertical acceleration of the ...

0

First of all, you mention that when the instrument disconnects, it is no longer rotating. Why would that be? However, the bigger issue is that if you want to conserve angular momentum, you have to measure it in the same way each time. For angular momentum, that means you have to retain the same axis each time. The instrument may no longer be rotating ...

1

Firstly let's look at the case of a horizontal plane first. In my answer here I derived that the critical friction coefficient is: $$\large{\mu_c=\frac{FI}{mg(I+mR^2)}}$$ Now we have three scenarios: a) No friction at all, $\mu=0$: Assuming no forces or couples act on the object then Newton tells us that the state of motion remains unchanged, or: ...

0

The moment of inertia is definitely affected by where the axis of rotation is located. To find the torque required to rotate an object where the axis of rotation is not through the center of mass, you definitely need to use the parallel axis theorem. If you intend to rotate a real-world object in such a fashion, expect a lot of "wobble" in the object if ...

0

In general, the position is the integral of the velocity - this is the essence of the equation you wrote down (without the "terminal velocity" part which makes no sense). If, at a certain time, you change the velocity (direction), you can just treat that point as the start of a new trajectory: the initial velocity will be "the velocity of the turned ...

0

Your issue is that a force $F$ applied through a distance $\Delta x$ results in more kinetic energy here than if the same force was applied through the same distance on a sliding frictionless block. The problem is the equation $W = F \Delta x$ only applies to point particles, or objects that move totally uniformly like point particles. When you have a ...

2

However, I find that the net torque of the cylinder is 0 if I set the axis of rotation as the bottom of the cylinder (the contact point between the ground and the cylinder); my first question is, what accounts for this discrepancy? I suspect there should be a force about the center of mass, but I don't see what could be the source of such force. You're ...

1

Your equations are not right: The moment of inertia about the pivot comes from the the mass at the top and by shifting the centre: $\frac{1}{2}MR^2+MR^2+M(2R)^2=\frac{11}{2}MR^2$ Now Apply the conservation of energy $\frac{1}{2}Iw^2=mg(2R+4R)$

1

Since the disk is pivoting around a point on its circumference (not its centre) you need to use the moment of inertia of the disk about that point, and it's not $1/2MR^2$.

1

Yes, rotational velocity and acceleration is shared by all points on a rigid body. We only state that a body rotated about a point because the linear velocity is zero at that point. See related answer here: http://physics.stackexchange.com/a/215165/392

0

The issue here is that there is a force acting on the axle to counter act gravity on that part of the disk, so not all parts of the disk will be accelerating uniformly at $9.81 \text{ m/s}^2$. Note also that this force acts at the axis of rotation, and therefore doesn't contribute to the torque. That is why you don't need to consider it when calculating ...

0

Background You already seem to know this stuff but it's worth going over again. So, the adjoint of an operator is the equivalent effect of the operator on the other side of the wavefunction inner product: $$\langle \Phi | \hat A | \Psi\rangle = \int_{-\infty}^{\infty} dx~\Phi^*(x) ~ A[\Psi](x) = \int_{-\infty}^{\infty} dx~{A^\dagger[\Phi]}^*(x) ~ \Psi(x) ... 0 Intuitively, shifting then reflecting is not the same as reflecting then shifting. Consider the case of first shifting 1 unit to the right from 0, then reflecting: you end up at x=-1. If you reflect first, it does nothing, and then shifting to the right by 1 means you end up at x=1. The problem is that$$\hat{T}(-a)\hat{R}\psi(x) =\hat{T}(-a)\psi(-x) ...

0

For convenience, I would first like to change the sign convention. That is, R_{z}(\alpha) = \begin{pmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{pmatrix},\quad R_{y}(\beta) = \begin{pmatrix}\cos\beta&0&\sin\beta\\0&1&0\\-\sin\beta&0&\cos\beta\end{pmatrix},\quad ...

0

Since the EM tensor is a tensor of rank two, the transformation requires two matrices: $$F^{\alpha\beta} \to F'^{\alpha\beta} = R^\alpha_\mu R^\beta_\nu F^{\mu\nu}$$ Or, in matrix form, $$F \to F' = RTR^T$$

2

The total energy at the top is $$T = \frac{1}{2} m v^2 + m g L$$ The total energy at some other point is $$B = \frac{1}{2} m f^2 + m g L \cos\theta$$ Energy is conserved so $$f = \sqrt{v^2 + 2 L g (1-\cos\theta) }$$

1

Physicists also use g/kg in many fields (chemistery, atmosphere dynamics), and Hubble constant is in (km/s)/Mpc. Light transport and scattering have many very close formula that might only differs by having or not a cosine inside, or be integrated or not with a unitary weight function. Of course you would be right to say g/kg is dimensionless and Hubble ...

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