Editing Angular momentum (section)

== Analogy to linear momentum ==
Angular momentum can be described as the rotational analog of [[linear momentum]]. Like linear momentum it involves elements of [[mass]] and [[Displacement (vector)|displacement]]. Unlike linear momentum it also involves elements of [[Position (vector)|position]] and [[shape]].

Many problems in physics involve matter in motion about some certain point in space, be it in actual rotation about it, or simply moving past it, where it is desired to know what effect the moving matter has on the point—can it exert energy upon it or perform work about it? [[Energy]], the ability to do [[Work (physics)|work]], can be stored in matter by setting it in motion—a combination of its [[inertia]] and its displacement. Inertia is measured by its [[mass]], and displacement by its [[velocity]]. Their product,
<math display="block">\begin{align}
(\text{amount of inertia}) \times (\text{amount of displacement})&=\text{amount of (inertia⋅displacement)}\\
\text{mass} \times \text{velocity} &= \text{momentum}\\
m \times v &= p\\
\end{align}
</math>
is the matter's [[momentum]].<ref>
{{cite book
 |url=https://books.google.com/books?id=bKEAAAAAMAAJ
 |title=Physics: Advanced Course
 |last=Barker
 |first=George F.
 |page=66
 |publisher=Henry Holt and Company, New York
 |edition=4th
 |date=1893
 |via=Google Books
}}</ref> Referring this momentum to a central point introduces a complication: the momentum is not applied to the point directly. For instance, a particle of matter at the outer edge of a wheel is, in effect, at the end of a [[lever]] of the same length as the wheel's radius, its momentum turning the lever about the center point. This imaginary lever is known as the ''moment arm''. It has the effect of multiplying the momentum's effort in proportion to its length, an effect known as a [[Moment (physics)|''moment''.]] Hence, the particle's momentum referred to a particular point,
<math display="block">\begin{align}
(\text{moment arm}) \times (\text{amount of inertia}) \times (\text{amount of displacement})&=\text{moment of (inertia⋅displacement)}\\
\text{length} \times \text{mass} \times \text{velocity} &= \text{moment of momentum}\\
r \times m \times v &= L\\
\end{align}
</math>
is the ''angular momentum'', sometimes called, as here, the ''moment of momentum'' of the particle versus that particular center point. The equation <math>L = rmv</math> combines a moment (a mass <math>m</math> turning moment arm <math>r</math>) with a linear (straight-line equivalent) speed <math>v</math>. Linear speed referred to the central point is simply the product of the distance <math>r</math> and the angular speed <math>\omega</math> versus the point: <math>v=r\omega,</math> another moment. Hence, angular momentum contains a double moment: <math>L = rmr \omega.</math> Simplifying slightly, <math>L = r^2 m\omega,</math> the quantity <math>r^2m</math> is the particle's [[moment of inertia]], sometimes called the second moment of mass. It is a measure of rotational inertia.<ref>
{{cite book
 |url=https://books.google.com/books?id=bKEAAAAAMAAJ
 |title=Physics: Advanced Course
 |last=Barker
 |first=George F.
 |pages=67–68
 |publisher=Henry Holt and Company, New York
 |edition=4th
 |date=1893
 |via=Google Books
}}</ref>

[[File:Moment of inertia examples.gif|thumb|right|[[Moment of inertia]] (shown here), and therefore angular momentum, is different for each shown configuration of [[mass]] and [[Rotation around a fixed axis|axis of rotation]].]]

The above analogy of the translational momentum and rotational momentum can be expressed in vector form:{{citation needed|reason=This recent addition appears to be OR, since it applied only in an overly simplified context.  The moment of inertial is not a scalar, but rather a matrix.|date=August 2022}}
* <math display="inline">\mathbf  p = m\mathbf  v</math> for linear motion
* <math>\mathbf  L = I\boldsymbol\omega</math> for rotation

The direction of momentum is related to the direction of the velocity for linear movement. The direction of angular momentum is related to the angular velocity of the rotation.

Because [[moment of inertia]] is a crucial part of the spin angular momentum, the latter necessarily includes all of the complications of the former, which is calculated by multiplying elementary bits of the mass by the [[Square (algebra)|squares]] of their [[distances]] from the center of rotation.<ref name="Oberg">
{{cite book
 |last1 = Oberg |first1 = Erik
 |display-authors=etal
 |title = Machinery's Handbook
 |edition=26th
 |publisher = Industrial Press, Inc., New York
 |isbn = 978-0-8311-2625-4
 |date=2000
 |page=143
}}</ref> Therefore, the total moment of inertia, and 
the angular momentum, is a complex function of the configuration of the [[matter]] about the center of rotation and the orientation of the rotation for the various bits.

For a [[rigid body]], for instance a wheel or an asteroid, the orientation of rotation is simply the position of the [[Rotation around a fixed axis|rotation axis]] versus the matter of the body. It may or may not pass through the [[center of mass]], or it may lie completely outside of the body. For the same body, angular momentum may take a different value for every possible axis about which rotation may take place.<ref>
{{cite book
 |last1 = Watson |first1 = W.
 |title = General Physics
 |publisher = Longmans, Green and Co., New York
 |date=1912
 |url=https://books.google.com/books?id=Yb8zAQAAMAAJ
 |page=34
 |via= Google books
}}</ref> It reaches a minimum when the axis passes through the center of mass.<ref>
{{cite book
 |last1 = Kent |first1 = William
 |title = The Mechanical Engineers' Pocket Book
 |edition=9th
 |publisher = John Wiley and Sons, Inc., New York
 |date=1916
 |url=https://books.google.com/books?id=DxoMAQAAIAAJ
 |page=517
 |via=Google books
}}</ref>

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. Their orientations may also be completely random.

In brief, the more mass and the farther it is from the center of rotation (the longer the [[Torque#Moment arm formula|moment arm]]), the greater the moment of inertia, and therefore the greater the angular momentum for a given [[angular velocity]]. In many cases the [[moment of inertia]], and hence the angular momentum, can be simplified by,<ref>
{{cite book
 |last1 = Oberg |first1 = Erik
 |display-authors=etal
 |title = Machinery's Handbook
 |edition=26th
 |publisher = Industrial Press, Inc., New York
 |isbn = 978-0-8311-2625-4
 |date=2000
 |page= 146
}}</ref>
<math display="block" qid=Q165618>I=k^2m,</math>where <math>k</math> is the [[radius of gyration]], the distance from the axis at which the entire mass <math>m</math> may be considered as concentrated.

Similarly, for a [[Point particle|point mass]] <math>m</math> the [[moment of inertia]] is defined as,
<math display="block">I=r^2m</math>where <math>r</math> is the [[radius]] of the point mass from the center of rotation, and for any collection of particles <math>m_i</math> as the sum,
<math display="block">\sum_i I_i = \sum_i r_i^2m_i .</math>

Angular momentum's dependence on position and shape is reflected in its [[Units of measurement|units]] versus linear momentum: kg⋅m<sup>2</sup>/s or N⋅m⋅s for angular momentum versus [[SI derived unit|kg⋅m/s]] or [[SI derived unit|N⋅s]] for linear momentum. When calculating angular momentum as the product of the moment of inertia times the angular velocity, the angular velocity must be expressed in radians per second, where the radian assumes the dimensionless value of unity. (When performing dimensional analysis, it may be productive to use [[Dimensional analysis#Siano's extension: orientational analysis|orientational analysis]] which treats radians as a base unit, but this is not done in the [[International system of units]]). The units if angular momentum can be interpreted as [[torque]]⋅time. An object with angular momentum of {{nowrap|''L'' N⋅m⋅s}} can be reduced to zero angular velocity by an angular [[Impulse (physics)|impulse]] of {{nowrap|''L'' N⋅m⋅s}}.<ref>
{{cite book
 |last1 = Oberg |first1 = Erik
 |title = Machinery's Handbook
 |edition=26th
 |publisher = Industrial Press, Inc., New York
 |isbn = 978-0-8311-2625-4
 |date=2000|display-authors=etal|pages=161–162
}}</ref><ref>
{{cite book
 |last1 = Kent |first1 = William
 |title = The Mechanical Engineers' Pocket Book
 |edition=9th
 |publisher = John Wiley and Sons, Inc., New York
 |date=1916
 |url=https://books.google.com/books?id=DxoMAQAAIAAJ
 |page=527
 |via= Google books
}}</ref>

The [[Plane (geometry)|plane]] [[perpendicular]] to the axis of angular momentum and passing through the center of mass<ref>
{{cite book
 |last = Battin
 |first = Richard H.
 |title = An Introduction to the Mathematics and Methods of Astrodynamics, Revised Edition
 |publisher = American Institute of Aeronautics and Astronautics, Inc.
 |isbn = 978-1-56347-342-5
 |date=1999}}, p. 97
</ref> is sometimes called the ''invariable plane'', because the direction of the axis remains fixed if only the interactions of the bodies within the system, free from outside influences, are considered.<ref name="Rankine">
{{cite book
 |last1=Rankine |first1=W. J. M.
 |title=A Manual of Applied Mechanics
 |edition=6th
 |publisher= Charles Griffin and Company, London
 |url=https://books.google.com/books?id=u9UKAQAAIAAJ
 |date=1872
 |page=507
 |via=Google books
}}</ref> One such plane is the [[Invariable plane|invariable plane of the Solar System]].

=== Angular momentum and torque ===
{{See also|Torque#Relationship with the angular momentum}}
[[Newton's laws of motion#Newton's second law|Newton's second law of motion]] can be expressed mathematically,
<math display="block">\mathbf{F} = m\mathbf{a},</math>
or [[force]] = [[mass]] × [[acceleration]]. The rotational equivalent for point particles may be derived as follows:
<math display="block">\mathbf{L} = I\boldsymbol{\omega}</math>
which means that the torque (i.e. the time [[derivative]] of the angular momentum) is
<math display="block" qid=Q48103>\boldsymbol{\tau} = \frac{dI}{dt}\boldsymbol{\omega} + I\frac{d\boldsymbol{\omega}}{dt}. </math>

Because the moment of inertia is <math>mr^2</math>, it follows that <math>\frac{dI}{dt} = 2mr\frac{dr}{dt} = 2rp_{||}</math>, and <math>\frac{d\mathbf{L}}{dt} = I\frac{d\boldsymbol{\omega}}{dt} + 2rp_{||}\boldsymbol{\omega},</math> which, reduces to
<math display="block">\boldsymbol{\tau} = I\boldsymbol{\alpha} + 2rp_{||}\boldsymbol{\omega}.</math>
This is the rotational analog of Newton's second law. Note that the torque is not necessarily proportional or parallel to the angular acceleration (as one might expect). The reason for this is that the moment of inertia of a particle can change with time, something that cannot occur for ordinary mass.