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
How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript
Examples
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Simple
a) Calculate Lrot (both magnitude and direction)
The rotational angular momentum is zero, because even though the apparatus is rotating clockwise with angular speed, the barbell maintains its vertical orientation, so the rotational angular momentum of the the barbell is zero. The angular velocity of the barbell is zero, so the rotational angular momentum of the barbell is also zero. The magnitude is zero and there is no direction for rotational angular momentum.
b) Calculate Ltrans,B.
Middling
Difficult
Connectedness
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Total angular momentum is very interesting to witness in the orbit of planets and satellites.
This image shows the translational angular momentum of the Earth (relative to the location of the Sun) and the rotational angular momentum (relative to the center of mass of the Earth).
It is also important to note that for point particle systems that there is no rotational angular momentum, and only translational angular momentum. This is important, because the total angular momentum for a point particle system is just simply translational angular momentum.
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 is also used for the Bohr model of the atom.
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See also
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