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Composition Setup
 Excerpt
The circulation of linear momentum about the specified axis, being proportional to the component of momentum or each mass along a circle about the axis and the radius of the circle. Angular momentum is changed by external torques, and therefore is constant when these sum to zero. The angular momentum of a rigid body is proportional to its moment of inertia times its angular velocity.
Motivation for Concept
Imagine a satellite spinning in space. While there is a gravitational force on the satellite which causes its linear momentum to change, the satellite will continue to spin at the same speed about its axis of rotation.  To change its rate of spin would require the application of an external force that did not act on the center of mass, but rather acted at some point off the axis of rotation. Moreover, the force would not change the spin rate if it acted directly toward or away from the axis - only if it had some component along or against the rotational velocity of the point where it was applied. 
For a rigid body, angular momentum is analogous to the linear momentum, but is a more complicated topic than conservation of linear momentum because internal motions of the system can change the rate of rotation of a system whereas internal motions cannot change its total linear momentum.  For example, if the satellite spread out solar panels, its rate of rotation would change, just as a twirling skater who pulls in his arms will rotate faster.  Thus angular velocity can change even when angular momentum about some point is conserved due to the fact that there is no applied torque about that axis.
Utility of Single-Axis Case
The specification of a single axis of rotation allows a complete description of the angular momentum only in certain cases. To ensure that a problem can be completely described using angular momentum about a single axis, the following conditions must hold.
Movement in a Plane: Rotational motion in introductory physics will be concerned only with systems where the position of the center of mass and the linear momentum of each constituent is confined to a plane (by our convention, the xy plane).
Angular Velocity and Angular Momentum Perpendicular to the Plane: Each rigid body in this system that is rotating must only rotate such that its angular velocity and angular acceleration are directed perpendicular to the xy plane containing the centers of mass, that is in the +z or -z direction.
...
Examples of Valid Systems
...
Systems that can be treated as having one-dimensional angular momentum include:
  • Ball rolling from rest down an inclined plane.
  • A yo-yo.
  • Non-spinning baseball moving horizontally strikes bat pivoted horizontally at the handle with ball center of mass at the same height as bat center of mass.
...
Examples of Invalid Systems
...
Systems that require more than 1-D angular momentum:
  • Styrofoam cup (wider on the top than the bottom) rolling down an inclined plane (the cup will begin to turn as it rolls).
  • Baseball moving horizontally with spin along a horizontal axis strikes bat pivoted horizontally at the handle with the ball center of mass at same height as bat center of mass (ball and bat angular momenta perpendicular).
  • Baseball moving horizontally with no spin strikes bat pivoted horizontally at the handle with ball center of mass above or below the bat center of mass (ball center of mass and bat center of mass move in different planes).
  • Gyroscopes (the one invalid system that is generally covered in introductory mechanics).
Conditions on the Chosen Axis of Rotation
For a consistent definition of angular momentum, the axis of rotation must satisfy one of two possibilities:
  1. The axis is fixed (does not move) in an inertial reference frame.
  2. The axis intersects the xy plane at the point of the system's center of mass. (Note that the system's center of mass may be moving and even accelerating.)
Definition of Angular Momentum
Angular Momentum of a Point Particle
Beginning with Newton's 2nd Law for a point particle:
 Latex
 Wiki Markup
{composition-setup}{composition-setup}
{table:border=1|frame=void|rules=cols|cellpadding=8|cellspacing=0}
{tr:valign=top}
{td:width=350|bgcolor=#F2F2F2}
{live-template:Left Column}
{td}
{td}

h1. Angular Momentum about a Single Axis

{excerpt}The circulation of [linear momentum|momentum] about the specified axis, being proportional to the component of momentum or each mass along a circle about the axis and the radius of the circle.  Angular momentum is changed by external [torques|torque (single-axis)], and therefore is constant when these sum to zero. The angular momentum of a rigid body is proportional to its [moment of inertia] times its angular velocity.{excerpt}

h3. Motivation for Concept

Imagine a satellite spinning in space. While there is a gravitational force on the satellite which causes its linear momentum to change, the satellite will continue to spin at the same speed about its axis of rotation.  To change its rate of spin would require the application of an external force that did not act on the center of mass, but rather acted at some point off the axis of rotation.  Moreover, the force would not change the spin rate if it acted directly toward or away from the axis - only if it had some component along or against the rotational velocity of the point where it was applied. 

For a rigid body, angular momentum is analogous to the linear momentum, but is a more complicated topic than conservation of linear momentum because internal motions of the system can change the rate of rotation of a system whereas internal motions cannot change its total linear momentum.  For example, if the satellite spread out solar panels, its rate of rotation would change, just as a twirling skater who pulls in his arms will rotate faster.  Thus angular velocity can change even when angular momentum about some point is conserved due to the fact that there is no applied torque about that axis.


h3. Utility of Single-Axis Case

The specification of a single [axis of rotation] allows a complete description of the angular momentum only in certain cases.   To ensure that a problem can be completely described using angular momentum about a single axis, the following conditions must hold.

h4. Movement in a Plane

Rotational motion in introductory physics will be concerned only with [systems|system] where the [position] of the [center of mass] and the linear [momentum] of each constituent is confined to a plane (by our convention, the _xy_ plane).

h4. Angular velocity and angular momentum Perpendicular to the Plane

Each rigid body in this system that is rotating must only rotate such that its angular velocity and angular acceleration are directed perpendicular to the _xy_ plane containing the centers of mass, that is in the \+z or \-z direction.

h4. Examples of Valid Systems

Systems that can be treated as having one-dimensional angular momentum include:
* Ball rolling from rest down an inclined plane.
* A yo-yo.
* Non-spinning baseball moving horizontally strikes bat pivoted horizontally at the handle with ball center of mass at the same height as bat center of mass.

h4. Examples of Invalid Systems

Systems that require more than 1-D angular momentum:
* Styrofoam cup (wider on the top than the bottom) rolling down an inclined plane (the cup will begin to turn as it rolls).
* Baseball moving horizontally with spin along a horizontal axis strikes bat pivoted horizontally at the handle with the ball center of mass at same height as bat center of mass (ball and bat angular momenta perpendicular).
* Baseball moving horizontally with no spin strikes bat pivoted horizontally at the handle with ball center of mass above or below the bat center of mass (ball center of mass and bat center of mass move in different planes).
* [Gyroscopes] (the one invalid system that is generally covered in introductory mechanics).

h3. Conditions on the Chosen Axis of Rotation

For a consistent definition of angular momentum, the [axis of rotation] must satisfy one of two possibilities:
# The axis is fixed (does not move) in an inertial reference frame.
# The axis intersects the _xy_ plane at the point of the *system's* center of mass.  (Note that the system's center of mass may be moving and even accelerating.)


h3. Definition of Angular Momentum


h4. Angular Momentum of a Point Particle

Beginning with [Newton's 2nd Law|Newton's Second Law] for a point particle:
{latex}\begin{large} \[ \sum \vec{F} = \frac{d(m\vec{v})}{dt} \] \end{large}{latex}

we 
...
 take 
...
 the 
...
cross 
...
 product
...
 of 
...
 the 
...
 particle's 
...
 position 
...
 measured 
...
 in 
...
 the 
...
xy
...
 plane 
...
 from 
...
 the 
...
 point 
...
 of 
...
 intersection 
...
 with 
...
 the 
...
 chosen 
...
axis 
...
 of 
...
 rotation
...
:

{
 Latex
}\begin{large}\[ \sum \vec{r}\times\vec{F} = \vec{r}\times \frac{d(m\vec{v})}{dt}\]\end{large}{latex}
Assuming _m_ is 
Assuming m is constant, 
...
 we 
...
 can 
...
 pass 
...
 it 
...
 out 
...
 of 
...
 the 
...
 derivative. 
...
 We 
...
 cannot 
...
 assume 
...
 that 
...
 it 
...
 is 
...
 reasonable 
...
 to 
...
 pass 
...
r
...
into
...
 the 
...
 derivative, 
...
 however, 
...
 since 
...
 it 
...
 certainly 
...
 has 
...
 a 
...
 time 
...
 dependence 
...
 (unless 
...
 the 
...
 velocity 
...
 is 
...
 zero). 
...
 We 
...
 can, 
...
 however, 
...
prove
...
 that 
...
 it 
...
 is 
...
 reasonable 
...
 to 
...
 pass 
...
r
...
 into 
...
 the 
...
 derivative. 
...
 To 
...
 see 
...
 this, 
...
 we 
...
 evaluate:

{
 Latex
}\begin{large} \[ \frac{d(\vec{r}\times\vec{v})}{dt} = \frac{d\vec{r}}{dt}\times \vec{v} + \vec{r}\times\frac{d\vec{v}}{dt} = \vec{v}\times\vec{v} + \vec{r}\times\frac{d\vec{v}}{dt}\]\end{large}{latex}

Noting 
...
 that 
...
 the 
...
 cross 
...
 product 
...
 of 
...
 any 
...
 vector 
...
 with 
...
 itself 
...
 is 
...
 zero, 
...
 we 
...
 then 
...
 see 
...
 that 
...
 if 
...
 the 
...
 particle's 
...
 mass 
...
 is 
...
 constant:

{
 Latex
}\begin{large} \[ \sum \vec{r}\times\vec{F} = \frac{d(m\vec{r}\times\vec{v})}{dt} \] \end{large}{latex}

The 
...
 left 
...
 hand 
...
 side 
...
 is 
...
 simply 
...
 the 
...
 net 
...
torque 
...
 (single-axis)
...
 on 
...
 the 
...
 particle 
...
 about 
...
 the 
...
 chosen 
...
 axis. 
...
 Thus, 
...
 in 
...
 the 
...
 absence 
...
 of 
...
 net 
...
 torque, 
...
 the 
...
 quantity:

{
 Latex
}\begin{large}\[ L_{z} = m\vec{r}\times\vec{v}\]\end{large}{latex}

is 
...
 constant 
...
 in 
...
 time. 
...
 We 
...
 therefore 
...
 choose 
...
 to 
...
 define 
...
 this 
...
 quantity 
...
 as 
...
 the 
...
 angular 
...
 momentum 
...
 of 
...
 the 
...
 point 
...
 particle 
...
 about 
...
 the 
...
 chosen 
...
 axis. 
...
 Note 
...
 that 
...
L
...
 is 
...
 a 
...
 vector 
...
 quantity, 
...
 but 
...
 it 
...
 is 
...
 one-dimensional 
...
 (it 
...
 must 
...
 point 
...
 perpendicular 
...
 to 
...
 the 
...
xy
...
 plane 
...
 because 
...
 of 
...
 the 
...
 cross 
...
 product 
...
 and 
...
 our 
...
 assumptions). 
...
 We 
...
 have 
...
 therefore 
...
 indicated 
...
 the 
...
 vector 
...
 nature 
...
 by 
...
 a 
...
 component 
...
 subscript 
...
 instead 
...
 of 
...
 a 
...
 vector 
...
 arrow.
...
Angular 
...
 Momentum 
...
 of 
...
 a 
...
 Rigid 
...
 Body 
...
 in 
...
 Pure 
...
 Rotation
...
By 
...
 dividing 
...
 a 
...
 rigid 
...
 body 
...
 into 
...
N
...
 tiny 
...
 mass 
...
 elements 
...
Δmi (i 
...
 ranging 
...
 from 
...
 1 
...
 to 
...
N
...
), 
...
 we 
...
 can 
...
 express 
...
 its 
...
 angular 
...
 momentum 
...
 about 
...
 a 
...
 specific 
...
 axis 
...
 as:

{
 Latex
}\begin{large}\[ L_{z} = \sum_{i=1}^{N} \vec{r}_{i}\times\vec{v}_{i} (\Delta m_{i}) \]\end{large}{latex}

The 
...
 assumption 
...
 of 
...
 pure 
...
 rotation, 
...
 however, 
...
 means 
...
 that 
...
 the 
...
 velocity 
...
 is 
...
 completely 
...
 tangential 
...
 (perpendicular 
...
 to 
...
r
...
) 
...
 and 
...
 equal 
...
 in 
...
 magnitude 
...
 to 
...
 the 
...
 magnitude 
...
 of 
...
r
...
ω 
...
 where 
...
 ω 
...
 is 
...
 the 
...
 angular 
...
 velocity 
...
 associated 
...
 with 
...
 the 
...
 rotation. 
...
 We 
...
 can 
...
 therefore 
...
 write:

{
 Latex
}\begin{large}\[ L_{z} = \sum_{i=1}^{N} r_{i}^{2} \omega \Delta m_{i} = \omega \sum_{i=1}^{N} r_{i}^{2} \Delta m_{i}\]\end{large}{latex}

where 
...
 we 
...
 have 
...
 removed 
...
 ω 
...
 from 
...
 the 
...
 sum 
...
 since 
...
 a 
...
 body 
...
 in 
...
 pure 
...
 rotation 
...
 has 
...
 a 
...
 uniform 
...
 angular 
...
 velocity.
...
In 
...
 the 
...
 limit 
...
 of 
...
 a 
...
 continuous 
...
 body, 
...
 the 
...
 sum 
...
 goes 
...
 over 
...
 to 
...
 an 
...
 integral:

{
 Latex
}\begin{large}\[ \sum r_{i}^{2} \Delta m_{i} \rightarrow \int r^{2} dm \]\end{large}{latex}

which 
...
 is 
...
 simply 
...
 the 
...
 definition 
...
 of 
...
 the 
...
 body's 
...
moment 
...
 of 
...
 inertia
...
 about 
...
 the 
...
 chosen 
...
 axis.
...
Thus, 
...
 for 
...
 a 
...
 rigid 
...
 body:

{
 Latex
}\begin{large}\[ L_{z} = I_{a} \omega\]\end{large}{latex}
where _I{_}{~}a~ is the }
where Ia is the body's 
...
 moment 
...
 of 
...
 inertia 
...
 about 
...
 the 
...
 chosen 
...
 axis.
...
Angular 
...
 Momentum 
...
 of 
...
 a 
...
 Rigid 
...
 Body 
...
 both 
...
 Translating 
...
 and 
...
 Rotating
...
A 
...
 rigid 
...
 body's 
...
 center 
...
 of 
...
 mass 
...
 can 
...
 be 
...
 translating 
...
 with 
...
 respect 
...
 to 
...
 the 
...
 axis 
...
 chosen. 
...
 If 
...
 we 
...
 again 
...
 imagine 
...
 dividing 
...
 the 
...
 body 
...
 into 
...
N
...
 mass 
...
 elements 
...
 Δmi, we note that these mass elements must execute pure rotation about the rigid body's center of mass.
 Note
If the mass elements of the rigid body could change their radial separation from the center of mass, then the relative distance between points in the body would change. Such a change would violate the definition of rigid body.
Because of this property of rigid bodies, it makes sense to restate the sum:
 Latex
{_}m{_}{~}i~, we note that these mass elements _must_ execute pure rotation about the rigid body's center of mass.
{note}If the mass elements of the rigid body could change their radial separation from the center of mass, then the relative distance between points in the body would change.  Such a change would violate the definition of [rigid body].
{note}
Because of this property of rigid bodies, it makes sense to restate the sum:
{latex}\begin{large}\[ L_{z} = \sum_{i=1}^{N} \vec{r}_{i}\times\vec{v}_{i} (\Delta m_{i}) \]\end{large}{latex}

by 
...
 defining 
...
 body 
...
 coordinates 
...
r
...
' 
...
 such 
...
 that:

{
 Latex
}\begin{large}\[ \vec{r} = \vec{r}\:' + \vec{r}_{cm}\]\end{large}{latex}
where _r{_}{~}cm~ is the position of the 
where rcm is the position of the body's 
...
 center 
...
 of 
...
 mass 
...
 measured 
...
 from 
...
 the 
...
 axis 
...
 of 
...
 rotation. 
...
 Taking 
...
 a 
...
 time 
...
 derivative 
...
 causes 
...
 us 
...
 to 
...
 similarly 
...
 define:

{
 Latex
}\begin{large}\[ \vec{v} = \vec{v}\:' + \vec{v}_{cm}\]\end{large}{latex}

We 
...
 can 
...
 now 
...
 use 
...
 the 
...
 constraint 
...
 of 
...
 the 
...
 rigid 
...
 body 
...
 (pure 
...
 rotation 
...
 about 
...
 the 
...
 center 
...
 of 
...
 mass) 
...
 to 
...
 define 
...
 an 
...
 ω
...
cm (the 
...
 angular 
...
 velocity 
...
 of 
...
 the 
...
 body 
...
 about 
...
 the 
...
 center 
...
 of 
...
 mass) 
...
 which 
...
 allows 
...
 us 
...
 to 
...
 write:

{
 Latex
}\begin{large}\[ \vec{r}\times\vec{v} = \omega_{cm}{r'}^{2} + \vec{r}_{cm}\times\vec{v}\:' + \vec{r}\:'\times\vec{v}_{cm}+\vec{r}_{cm}\times\vec{v}_{cm}\]\end{large}{latex}

Inserting 
...
 this 
...
 expression 
...
 into 
...
 the 
...
 sum 
...
 for 
...
 the 
...
 body's 
...
 angular 
...
 momentum 
...
 gives:

{
 Latex
}\begin{large}\[ L_{z} = \sum_{i=1}^{N} (\omega_{cm}{r'}_{i}^{2} + \vec{r}_{cm}\times\vec{v}_{i}\:' + \vec{r}_{i}\:'\times\vec{v}_{cm} + \vec{r}_{cm}\times\vec{v}_{cm}) (\Delta m_{i}) \]\end{large}{latexlarge}

In 
...
 the 
...
 primed 
...
 coordinates, 
...
 however, 
...
 the 
...
 center 
...
 of 
...
 mass 
...
 velocity 
...
 and 
...
 the 
...
 center 
...
 of 
...
 mass 
...
 position 
...
 are 
...
by 
...
 definition
...
 equal 
...
 to 
...
 zero. 
...
 Thus:

{
 Latex
}\begin{large} \[ \sum_{i=1}^{N} \vec{v}_{i}\:' \:\Delta m_{i} =\sum_{i=1}^{N} \vec{r}_{i}\:'\:\Delta m_{i} = 0 \]\end{large}{latex}

Therefore, 
...
 we 
...
 have:

{
 Latex
}\begin{large} \[ L_{z} = \left(\omega_{cm}\sum_{i=1}^{N} {r'}_{i}^{2} \Delta m_{i}\right) + \left(\vec{r}_{cm}\times\vec{v}_{cm}\right) \left(\sum_{i=1}^{N} \Delta m_{i}\right) = I_{cm}\omega_{cm} + M_{tot} \vec{r}_{cm}\times\vec{v}_{cm}\]\end{large}{latex}
where _M{_}{~}tot~ is the total mass of the rigid body.

Comparing this expression to those found in the previous sections, we see that the angular momentum of a rigid body that is both translating and rotating with respect to the chosen axis of rotation can be thought of as the sum of the angular momentum of the center of mass treated as a point particle with the mass of the entire rigid body and the angular momentum contributed by the body's rotation about the center of mass.

h3. Advanced Section:  Valid Choices for the Axis of Rotation

Under construction
where Mtot is the total mass of the rigid body.
Comparing this expression to those found in the previous sections, we see that the angular momentum of a rigid body that is both translating and rotating with respect to the chosen axis of rotation can be thought of as the sum of the angular momentum of the center of mass treated as a point particle with the mass of the entire rigid body and the angular momentum contributed by the body's rotation about the center of mass.