# What is the difference between translational angular momentum and rotational angular momentum?

We were taught that total angular momentum of a body equals the translational angular momentum plus the rotational angular momentum: $$\boldsymbol L=\boldsymbol r\times\boldsymbol p+I\boldsymbol \omega$$ If so, how?

• "Translational" and "rotational" angular momentum are not really standard terms, I should think. The concepts of extrinsic and intrinsic angular momentum, which do get standard use, are probably better descriptors of what you actually mean. – Emilio Pisanty Oct 24 '17 at 17:48
• What do you mean by extrinsic and intrinsic angular momentum? – Garry Host Oct 24 '17 at 18:11
• – ja72 Oct 25 '17 at 2:58
• -1. Unclear. Please can you explain your difficulty. – sammy gerbil Oct 25 '17 at 12:34

Here is a solution I came up with:

By definition: $$\vec{L}=\int{\vec{r}\times d\vec{p}}=\int{\vec{r}\times\vec{v}\ dm}$$ where integration is carried out over the whole mass of the body.

Now we can write $\vec{r}$ as $\vec{r}_G+\vec{r}_R$ where $\vec{r}_G$ is the position vector of the center of mass and $\vec{r}_R$ is the position vector of our mass infinitesimal in the center of mass frame. Similarly, we can write $\vec{v}$ as $\vec{v}_G+\vec{v}_R$ where as above, $\vec{v}_G$ is velocity of the center of mass and $\vec{v}_R$ is the velocity of the mass infinitesimal in the center of mass frame. The equation becomes:$$\vec{L}=\int{(\vec{r}_G+\vec{r}_R)\times(\vec{v}_G+\vec{v}_R)\ dm}$$ We can expand that:$$\vec{L}=\int{\vec{r}_G\times \vec{v}_G\ dm}+\int{\vec{r}_G\times \vec{v}_R\ dm}+\int{\vec{r}_R\times \vec{v}_G\ dm}+\int{\vec{r}_R\times \vec{v}_R\ dm}$$The first term in the LHS is simply the $\vec{r}\times \vec p$ term in your equation; it is the angular momentum in case all the mass was concentrated at the center of mass. The last term in the LHS is the $I\vec w$ term in your equation; the angular momentum in the center of mass frame. As for the second term:$$\int{\vec{r}_G\times \vec{v}_R\ dm}=\vec r_G\times \int{\vec v_R\ dm}=\vec r_G\times \vec 0=\vec 0$$ since there is no net motion in the center of mass frame. Same for the third term: $$\int{\vec{r}_R\times \vec{v}_G\ dm}=\int{\vec{r}_R\ dm}\ \times\vec v_G=\vec 0\times \vec{v}_G=\vec 0$$since the mean position of the mass is the position of the center of mass; the zero vector in the center of mass frame. Thus we find your equation:$$\vec L=\vec r\times \vec p+I\vec w$$

This formula is useful when an object rotates about a point other than its center of mass. One could calculate the moment of inertia around this point, but that is not always convenient. Instead, one can decompose it into the angular momentum associated with the center of mass rotating around the center of rotation (what you called the "translational angular momentum") and angular momentum associated with the rotation of the object around its center of mass (what you called "rotational angular momentum"). I don't know whether those are common terms or not, but its the first I've heard of them.

The equation deriving this can be found on Wikipedias page on angular momentum under Collections of Particles.

The two parts are easily explained.

• First is the moment of momentum $\mathbf{r}_C \times \mathbf{p}$ because the line of action of momentum is away from the point of angular momentum measurement. The is the angular momentum because the center of mass is moving.

• The second part ${\rm I}_C\, {\boldsymbol \omega}$ is the angular momentum of the rotation about the center of mass. Since the object is not concentrated at single point this includes the sum of all the momentum vectors of each particle as it rotates about the center of mass.

To fully understand this relationship consider what the linear and angular momentum are about a point A not at the center of mass C.

\begin{align} \mathbf{p} & = m \mathbf{v}_C = m \left( \mathbf{v}_A + {\boldsymbol \omega} \times \mathbf{r}_C \right) \\ \mathbf{L}_A & = \mathbf{L}_C + \mathbf{r}_C \times \mathbf{p} = {\rm I}_C {\boldsymbol \omega} + m\, \mathbf{r}_C \times \left( \mathbf{v}_A + {\boldsymbol \omega} \times \mathbf{r}_C \right) \end{align}

Conceptually this is grouped together as a big 6×6 system of equations, in block form:

$$\begin{Bmatrix} \mathbf{p} \\ \mathbf{L}_A\end{Bmatrix} = \begin{bmatrix} m\, {\boldsymbol 1} & -m \mathbf{r}_C \times \\ m \mathbf{r}_C \times & {\rm I}_C-m \mathbf{r}_C \times \mathbf{r}_C \times \end{bmatrix} \begin{Bmatrix} \mathbf{v}_A \\ {\boldsymbol \omega} \end{Bmatrix}$$

Use 3×3 the skew-symmetric cross product operator $\mathbf{r}_C \times = \left[\matrix{0&-z&y \\ z &0&-x \\ -y&x& 0}\right]$ and the expression on the bottom right of the big 6×6 matrix is the parallel axis theorem in vector form. It represents the effective mass moment of inertia of the body about A.

So your original question is really a question of why the $m \,\mathbf{r}_C\times$ cross terms on the upper right and bottom left? The reason those are there is because the axis the momentum goes through and the rotation axis of a rigid body are always at a distance from each other. As a result, the rotation away from the center of mass moves the center of mass and causes the term on the upper right, and symmetrically the lower left term.

That is, the moment of momentum $\mathbf{r}_C \times (m \mathbf{v}_A)$ adds to the angular momentum when transforming from point C to point A, and the term $m \left(-{\boldsymbol \omega} \times \mathbf{r}_C \right)$ adds to linear momentum when transforming from point C to point A also.

So the cross terms are due to a transformation of coordinates, and when the point of interest is at the center of mass with $\mathbf{r}_C = 0$ the above 6×6 matrix becomes block diagonal and the cross terms are zero.

$$\begin{Bmatrix} \mathbf{p} \\ \mathbf{L}_C\end{Bmatrix} = \begin{bmatrix} m\, {\boldsymbol 1} & 0 \\ 0 & {\rm I}_C \end{bmatrix} \begin{Bmatrix} \mathbf{v}_C \\ {\boldsymbol \omega} \end{Bmatrix}$$