Total Angular Momentum
Work in Progress by Fatima Jamil
Total angular momentum can be expressed as 
. This page explains the breakdown of total angular momentum in these 2 components to help understand the difference between rotational angular momentum and translational angular momentum.
The Main Idea
It is conveniant to break apart total angular momentum for a multiparticle system into rotational angular momentum and translational angular momentum. The translational angular momentum is associated with a rotation of the center of mass about some point A. This differs for different choices of the location of point A. The rotational angular momentum is associated with a rotation about the center of mass. The rotational angular momentum is independent of the location of the point A and the motion of the center of mass.
A Mathematical Model
It is often convenient to consider the angular momentum of a collection of particles about their center of mass, because this simplifies the mathematics considerably. The angular momentum of a collection of particles is the sum of the angular momenta of each particle:
[math]\displaystyle{ \mathbf{L}=\sum_i \mathbf{R}_i\times m_i \mathbf{V}_i }[/math] where [math]\displaystyle{ R_i }[/math] is the distance of particle i from the reference point, [math]\displaystyle{ m_i }[/math] is its mass, and [math]\displaystyle{ V_i }[/math] is its velocity. The center of mass is defined by:
[math]\displaystyle{ \mathbf{R}=\frac{1}{M}\sum_i m_i \mathbf{R}_i }[/math] where [math]\displaystyle{ M }[/math] is the total mass of all the particles.
If we define [math]\displaystyle{ \mathbf{r}_i }[/math] as the displacement of particle i from the center of mass, and [math]\displaystyle{ \mathbf{v}_i }[/math] as the velocity of particle i with respect to the center of mass, then we have
[math]\displaystyle{ \mathbf{R}_i=\mathbf{R}+\mathbf{r}_i\, }[/math] and [math]\displaystyle{ \mathbf{V}_i=\mathbf{V}+\mathbf{v}_i\, }[/math] In this case, the total angular momentum is:
[math]\displaystyle{ \mathbf{L}=\sum_i (\mathbf{R}+\mathbf{r}_i)\times m_i (\mathbf{V}+\mathbf{v}_i) = \left(\mathbf{R}\times M\mathbf{V}\right) + \left(\sum_i \mathbf{r}_i\times m_i \mathbf{v}_i\right) }[/math]
A Computational Model
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Understanding the topic of total angular momentum is extremely important in applying the conservation of angular momentum. Due to the fact that angular momentum is conserved, then it is important to note that if there is net force on some body directed towards a fix point, the center, then there is no torque on the body with respect to the center, and the angular momentum of the body about the center is constant. There is a very useful application of constant angular momentum, specifically seen in dealing with the orbits of planets and satellites. This concept
The conservation of angular momentum is used extensively in analyzing what is called central force motion. If the net force on some body is always directed toward a fixed point, the center, then there is no torque on the body with respect to the center, and the angular momentum of the body about the center is constant. Constant angular momentum is extremely useful when dealing with the orbits of planets and satellites. This concept was also used for the Bohr model of the atom.
The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation (or close to her body). By bringing part of her body mass closer to the axis, she decreases her body's moment of inertia. Because angular momentum is constant in the absence of external torques, the angular velocity (rotational speed) of the skater has to increase.
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