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This looks to me like a Jaynes-Cummings Hamiltonian, describing the interaction of a two-level atom with a quantized light field. For one polarization of the light field (k=1) the total Hamiltonian is made up of the atoms hamiltonian, the field hamiltonian and the interaction hamiltonian $$H_{atom} = \frac{\hbar\omega_a}{2}\sigma_z\\ H_{field} = \hbar\... -1 There is a very quick and clean way of doing this, which is presented in Building an Orthonormal Basis from a 3D Unit Vector Without Normalization. JR Frisvad. J. Graphics Tools 16 no. 3, 151 (2012). Suppose you have a normalized vector \vec n=(n_x,n_y,n_z)^T, and you want a rotation matrix that will take the z axis into \vec n. (Here it's ... 0 I think the original post was asking what a stellar day is as the sidereal day is a rotation of the earth and a solar day is a sidereal day + the missing 3 minutes 56 sec needed to = 24 hours. So why do we need a stellar day and what is it used for? To my understanding Stellar day is the more accurate name of a full 26,000 years rotations as a sidereal day ... 1 Since the action of a unitary matrix U on the Bloch sphere is defined by$$\sigma_j\mapsto U\,\sigma_j\,U^\dagger = U\,\sigma_j\,U^{-1}\tag{1} with $\sigma_j$ standing for the Pauli matrices, any scalar phase factor $e^{i\,\phi}$ is quotiented away by the map (which is the big A Adjoint representation of the unitary group). Unitary matrices always act ...

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No. As you realise, this will increase the moment of inertia, which will reduce angular acceleration since the torque which the motor can supply is limited. However, the maximum speed which the motor can reach is not affected, since this depends (mainly) on the aerodynamic force, which is the same - it depends on the shape and size of fan, but not its mass....

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There is nothing wrong with your proposed approach, but I think maybe you may have misunderstood your physics teacher; most likely he/ she would like you to be completely comfortable and proficient with the analysis of dynamical systems from an inertial frame (i.e. unaccelerated frame) before shifting on to analysis from accelerated frames, which involve ...

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Proceed by choosing a point O on the symmetry axis Z. Choose a differential fluid element having dimensions $\mathrm{d} r$, $\mathrm{d}x$ ,$\mathrm{d}y$ where x is the polar angle in your horizontal plane of observation and y is the azimuthal angle perpendicular to this plane. An F.B.D of this element should easily give the following 2 equations: \$P_z = p -...

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If you impose a non slip condition, then you say that during the collision both disks have the same tangential speed. You can calculate this speed (both the translational speed and the rotational speed contribute) and impose a final common tangential speed, plus the conservation of the angular momentum. The problem with this is that now you have an ...

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It is true that the weight will pull the machine to the north when it points north because the machine is pulling the weight to the south; by Newton's Third Law, that means the weight is pulling the machine northward. However, you have to think about what happens during the rest of the rotation. To have a faster velocity while the weight points north, the ...

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I think you have a small misconception about rotation sensing. Accelerometers will sense a frame's rotation of constant angular velocity as well as rotation where the angular velocity varies. This is because constant angular velocity rotation is itself an acceleration: something must undergo a centripetal acceleration to follow a curved path at constant ...

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