# angular momentum dimensional formula

Similarly, for a point mass {\displaystyle L=rmr\omega .}

{\displaystyle \mathbf {L} (\mathbf {r} ,t)} Equivalently, in Hamiltonian mechanics the Hamiltonian can be described as a function of the angular momentum.  However, Hayward's article apparently was the first use of the term and the concept seen by much of the English-speaking world.

2 But once you start pedalling, these wheels pick up angular momentum. {\displaystyle \mathbf {L} } )

, p {\displaystyle m_{i}} 2 As a planet orbits the Sun, the line between the Sun and the planet sweeps out equal areas in equal intervals of time. θ {\displaystyle L_{z}} Note, that for combining all axes together, we write the kinetic energy as: where pr is the momentum in the radial direction, and the moment of inertia is a 3-dimensional matrix; bold letters stand for 3-dimensional vectors. {\displaystyle \mathbf {F} } However, if the particle's trajectory lies in a single plane, it is sufficient to discard the vector nature of angular momentum, and treat it as a scalar (more precisely, a pseudoscalar). Quantum particles do possess a type of non-orbital angular momentum called "spin", but this angular momentum does not correspond to actual physical spinning motion..  Spin is often depicted as a particle literally spinning around an axis, but this is a misleading and inaccurate picture: spin is an intrinsic property of a particle, unrelated to any sort of motion in space and fundamentally different from orbital angular momentum. where r and for any collection of particles = Here the angular momentum is given by: Extended object: The object, which is rotating about a fixed point. The conservation of angular momentum is used in analyzing central force motion. Their orientations may also be completely random. m ( = R As an example, consider decreasing of the moment of inertia, e.g. It is an important quantity in physics because it is a conserved quantity—the total angular momentum of a closed system remains constant.

V L i i r 2

which, reduces to.   the quantity ⁡ i   about the axis

{\displaystyle mr^{2}} Expanding For a collection of objects revolving about a center, for instance all of the bodies of the Solar System, the orientations may be somewhat organized, as is the Solar System, with most of the bodies' axes lying close to the system's axis.

Thus, assuming the potential energy does not depend on ωz (this assumption may fail for electromagnetic systems), we have the angular momentum of the i-th object: We have thus far rotated each object by a separate angle; we may also define an overall angle θz by which we rotate the whole system, thus rotating also each object around the z-axis, and have the overall angular momentum: From Euler-Lagrange equations it then follows that: Since the lagrangian is dependent upon the angles of the object only through the potential, we have: Suppose the system is invariant to rotations, so that the potential is independent of an overall rotation by the angle θz (thus it may depend on the angles of objects only through their differences, in the form i The fact that the physics of a system is unchanged if it is rotated by any angle about an axis implies that angular momentum is conserved..

{\displaystyle \mathbf {L} =\sum _{i}\left(\mathbf {R} _{i}\times m_{i}\mathbf {V} _{i}\right)} , Bernoulli wrote in a 1744 letter of a "moment of rotational motion", possibly the first conception of angular momentum as we now understand it.. m i i

(For one particle, J = L + S.) Conservation of angular momentum applies to J, but not to L or S; for example, the spin–orbit interaction allows angular momentum to transfer back and forth between L and S, with the total remaining constant.

i Q8: If the moment of inertia of an isolated system is halved. ω This simple analysis can also apply to non-circular motion if only the component of the motion which is perpendicular to the radius vector is considered. M Instead of a steady pattern of rotation, you will observe something more complicated, with the shoe changing its orientation with respect to the rotation axis. Note that r The centripetal force on this point, keeping the circular motion, is: Thus the work required for moving this point to a distance dz farther from the center of motion is: For a non-pointlike body one must integrate over this, with m replaced by the mass density per unit z. The typical value ranges from 0 to 1. L ) {\displaystyle r}

=

and reducing, angular momentum can also be expressed, where In classical mechanics, the angular momentum of a particle can be reinterpreted as a plane element: in which the exterior product ∧ replaces the cross product × (these products have similar characteristics but are nonequivalent).
{\displaystyle \mathbf {L} =rmv\mathbf {\hat {u}} } r ^ It is a measure of rotational inertia..

{\displaystyle m}

ω

The interplay with quantum mechanics is discussed further in the article on canonical commutation relations.

|   and the linear momentum z {\displaystyle L=rmv\sin(\theta ),} {\displaystyle {\frac {dI}{dt}}=2mr{\frac {dr}{dt}}=2rp_{||}}

i

i

This imaginary lever is known as the moment arm. 2 {\displaystyle \mathbf {p} } L

Instead, the momentum that is physical, the so-called kinetic momentum (used throughout this article), is (in SI units), where e is the electric charge of the particle and A the magnetic vector potential of the electromagnetic field.

The total angular momentum of the collection of particles is the sum of the angular momentum of each particle, L

v r By the definition of the cross product, the is the rotation operator that takes any system and rotates it by angle {\displaystyle p_{x}}

{\displaystyle {\cal {L}}}

r r  Note, however, that this is no longer true in quantum mechanics, due to the existence of particle spin, which is angular momentum that cannot be described by the cumulative effect of point-like motions in space. i Newton derived a unique geometric proof, and went on to show that the attractive force of the Sun's gravity was the cause of all of Kepler's laws. The net external torque on any system is always equal to the total torque on the system; in other words, the sum of all internal torques of any system is always 0 (this is the rotational analogue of Newton's Third Law). I {\displaystyle =r\omega } ,