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Well, OK, this is a resolutely vague question, but there is something special, actually, about angular momentum dimensions. In quantum mechanics, the fundamental constant, $\hbar$ has dimensions of angular momentum (and is very small in terms of angular momenta, actions, or phase-space areas of our macroscopic world experience). Classical ...
We may say that $\hbar/2$ is the elementary quantum of the angular momentum. (The orbital angular momentum is a multiple of $\hbar$ without the factor of 1/2.) Just with some knowledge of Noether's theorem that links conservation laws and symmetries, one could have been able to guess – before he learned the full quantum mechanics – that the ...
The unit for angular acceleration α α is: rad/s2 r a d / s 2. The unit for torque is Nm N m: kg m2/s2 k g m 2 / s 2. And their relationship with Inertia is: I = τ/α I = τ / α. So shouldn't the unit for for Inertia be: kg m2/rad k g m 2 / r a d. yet everywhere I read says it is simply kg m2 k g m 2 instead.
Dimensionally, 1 radian is just that same as the pure number 1, but sometimes it is useful to distinguish between them. However, that distinction cannot be usefully maintained in more complicated calculations, so it is not really meaningful or useful to try to distinguish, say, the units of action from the units of angular momentum.
First is the moment of momentum $\mathbf{r}_C \times \mathbf{p}$ because the line of action of momentum is away from the point of angular momentum measurement. The is the angular momentum because the center of mass is moving. The second part ${\rm I}_C\, {\boldsymbol \omega}$ is the angular momentum of the rotation about the center of mass.
5. The angular momentum of the nucleus is the combined contribution of the spin-orbit angular momenta of the constituent particles. In order for an entity to have orbital angular momentum of its own it must some conceptual orbit: electrons in the atom, protons and neutrons in the nucleus, atoms in a molecule.
The proton and neutron are both fermions, and a state containing two of them must be antisymmetric under exchange. If they are bound without any orbital angular momentum, the only way to make the state antisymmetric is for the two particles to occupy a spin singlet. So the ground state of the diproton or dineutron must have spin zero.
According to multiple sources the SI units for angular momentum are kg * m$^2$ / sec.
0. Yes, angular momentum is fundamental, even in classical physics. To understand this, consider the Helmholtz decomposition of momentum density (p = ρu p = ρ u, where ρ ρ is inertial density and u u is velocity): p = p0 + ∇Φ + 1 2∇ ×s p = p 0 + ∇ Φ + 1 2 ∇ × s.
The moment of (linear) momentum, $\vec r \times \vec p$, where $\vec p = m\vec v$ is typically called angular momentum. The moment of a force, $\vec r \times \vec F$, is typically called torque. The first moment of mass of a collection of particles, $\sum_i m_i \vec x_i$, is the total mass times the center of mass position.