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{excerpt}The rotational equivalent of [linear momentum|momentum].  Angular momentum is often approximately conserved in collisions, and is usually conserved when external [torques|torque] sum to zero.  The action of purely internal forces can sometimes create a change in the system [moment of inertia] while conserving the angular momentum, which leads to interesting effects.{excerpt}

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h2. Motivation for Concept

Imagine a satellite spinning in space.  In the absence of external influences, the satellite will continue to spin.  To stop the spinning would require the application of force.  The faster the satellite is spinning, the more work the force will have to perform to halt the satellite.  These concepts are completely analogous to the linear ideas of force and work.  As such, there must be some quantity analogous to the linear momentum which is conserved in the absence of external influences.  Angular momentum is that quantity.  

Conservation of angular momentum is a more complicated topic than conservation of linear momentum, however.  It is very rare for an everyday system to experience a significant change in mass.  Thus, in physics problems involving conservation of linear momentum, the center of mass velocity of the system is usually constant.  It is not difficult, however, for a system to alter its moment of inertia.  In a case where angular momentum is conserved, such an alteration must produce an inverse alteration in the rotational speed.  This is easily seen in sporting events like figure skating and diving.  When an athlete is rotating essentially free of external forces (such as a figure skater in a spin on slippery ice or a diver rotating in mid-air) they can still affect a dramatic change in their rotational rate simply by tucking in their arms and legs or, conversely, by extending their arms and legs.    

h2. Conditions for One-Dimensional Angular Momentum

With the exception of certain specific systems like the [gyroscope], angular momentum problems in introductory physics will be confined to one dimension.  To ensure that a problem can be treated using 1-D angular momentum, the following conditions must hold.

h4. Movement in a Plane

Rotational motion in introductory physics will be concerned only with systems where the center of mass of each constituent is confined to move in a plane (we will refer to this as the _xy_ plane).  

h4. Rotation Perpendicular to the Plane

Each rigid body in the system that is rotating must only rotate such that their angular velocity and angular acceleration are directed perpendicular to the _xy_ plane containing the centers of mass.  (They must point in the +z or -z direction.)

h4. Rotational Symmetry
{anchor:rotational symmetry}

Finally, each [rigid body] in the system must obey *one* of the following rotational symmetry constraints:

# The object has a rotational symmetry about the axis of rotation.
# The object has a reflection symmetry about the _xy_ plane.

{note}Each class of objects in the table of common [moments of inertia|moment of inertia#table] _and their [parallel axis theorem] tranformations_ possess the second symmetry (some also possess the first).{note}

h4. 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. 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).


h2. Definition Conditions on the Chosen Axis of Angular MomentumRotation

h4. Angular MomentumFor a consistent definition of aangular 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 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 fromat the point of intersection with the *system's* center of mass.  (Note that the chosen [axissystem's center 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 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 evaluatemass may be moving and even accelerating.)


h2. 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} \[ \frac{d(sum \vec{r}\times\vec{v})}{dt} =F} = \vec{r}\times \frac{d(m\vec{rv})}{dt}\times \vec{v} + \vec{r}\times\frac{d\vec{v}}{dt} = \vec{v}\times\vec{v} + \vec{r}\times\frac{d\vec{v}}{dt}\]\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] on the particle about the chosen axis.  Thus, in the absence of net torque, the quantity]\end{large}{latex}

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} \[ L_{z} = m\frac{d(\vec{r}\times\vec{v})}{dt}\]\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.

h4.  Angular Momentum of a Rigid Body in Pure Rotation

By dividing a rigid body into _N_ tiny mass elements Δ_m_~i~ (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}\[ L_{z} =  = \frac{d\vec{r}}{dt}\times \vec{v} + \vec{r}\times\frac{d\vec{v}}{dt} = \vec{v}\times\vec{v} + \vec{r}\times\frac{d\vec{v}}{dt}\]\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] 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.

h4.  Angular Momentum of a Rigid Body in Pure Rotation

By dividing a rigid body into _N_ tiny mass elements Δ_m_~i~ (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 body's moment of inertia about the chosen axis.

h4. 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 Δ_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 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}{latex}

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} \[ L_{z} = \left(\omega_{cm}\sum_{i=1}^{N} r'_{i}^{2} \Delta m_{i}\right) + \left(\vec{r}_{cm}\times\vec{v}_{cm}\right) \left(\sum_{i=1}^{N} r_{i}^{2} \omega \Delta m_{i}\right) = I_{cm}\omega \sum_{i=1}^{N} r_{i}^{2} \Delta m_{icm} + M_{tot} \vec{r}_{cm}\times\vec{v}_{cm}\]\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 body's moment of inertia about the chosen axis.

h4. Angular Momentum of a Rigid Body both Translating and Rotating

_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.