Solid bodies

Angular momentum is also an extremely useful concept for describing rotating rigid bodies such as a gyroscope or a rocky planet. For a continuous mass distribution with density function ρ(r), a differential volume element dV with position vector r within the mass has a mass element dm = ρ(r)dV. Therefore, the infinitesimal angular momentum of this element is:

${\displaystyle d\mathbf{L}=\mathbf{r}\times <!--\; \times \; -->dm\mathbf{v}=\mathbf{r}\times <!--\; \times \; -->\mathrm{\rho <!--\; \rho \; -->}(\mathbf{r})dV\mathbf{v}=dV\mathbf{r}\times <!--\; \times \; -->\mathrm{\rho <!--\; \rho \; -->}(\mathbf{r})\mathbf{v}}$

and integrating this differential over the volume of the entire mass gives its total angular momentum:

${\displaystyle \mathbf{L}={\int <!--\; \int \; -->}_{V}dV\mathbf{r}\times <!--\; \times \; -->\mathrm{\rho <!--\; \rho \; -->}(\mathbf{r})\mathbf{v}}$

In the derivation which follows, integrals similar to this can replace the sums for the case of continuous mass.

Collection of particlesedit

For a collection of particles in motion about an arbitrary origin, it is informative to develop the equation of angular momentum by resolving their motion into components about their own center of mass and about the origin. Given,

${\displaystyle m_i}$ is the mass of particle ${\displaystyle i}$,

${\displaystyle {\mathbf{R}}_i}$ is the position vector of particle ${\displaystyle i}$ vs the origin,

${\displaystyle {\mathbf{V}}_i}$ is the velocity of particle ${\displaystyle i}$ vs the origin,

${\displaystyle \mathbf{R}}$ is the position vector of the center of mass vs the origin,

${\displaystyle \mathbf{V}}$ is the velocity of the center of mass vs the origin,

${\displaystyle {\mathbf{r}}_i}$ is the position vector of particle ${\displaystyle i}$ vs the center of mass,

${\displaystyle {\mathbf{v}}_i}$ is the velocity of particle ${\displaystyle i}$ vs the center of mass,

The total mass of the particles is simply their sum,

${\displaystyle M=\underset{i}{\sum <!--\; \sum \; -->}{m}_i.}$

The position vector of the center of mass is defined by,

${\displaystyle M\mathbf{R}=\underset{i}{\sum <!--\; \sum \; -->}{m}_i{\mathbf{R}}_i.}$

By inspection,

${\displaystyle {\mathbf{R}}_i=\mathbf{R}+{\mathbf{r}}_i}$ and ${\displaystyle {\mathbf{V}}_i=\mathbf{V}+{\mathbf{v}}_i.}$

The total angular momentum of the collection of particles is the sum of the angular momentum of each particle,

Expanding ${\displaystyle {\mathbf{R}}_i}$,

Expanding ${\displaystyle {\mathbf{V}}_i}$,

It can be shown that (see sidebar),

Prove that ${\displaystyle \underset{i}{\sum <!--\; \sum \; -->}{m}_i{\mathbf{r}}_i=\mathbf{0}}$ which, by the definition of the center of mass, is ${\displaystyle \mathbf{0},}$ and similarly for ${\displaystyle \underset{i}{\sum <!--\; \sum \; -->}{m}_i{\mathbf{v}}_i.}$

${\displaystyle \underset{i}{\sum <!--\; \sum \; -->}{m}_i{\mathbf{r}}_i=\mathbf{0}}$ and ${\displaystyle \underset{i}{\sum <!--\; \sum \; -->}{m}_i{\mathbf{v}}_i=\mathbf{0},}$

therefore the second and third terms vanish,

${\displaystyle \mathbf{L}=\underset{i}{\sum <!--\; \sum \; -->}\mathbf{R}\times <!--\; \times \; -->m_i\mathbf{V}+\underset{i}{\sum <!--\; \sum \; -->}{\mathbf{r}}_i\times <!--\; \times \; -->m_i{\mathbf{v}}_i.}$

The first term can be rearranged,

${\displaystyle \underset{i}{\sum <!--\; \sum \; -->}\mathbf{R}\times <!--\; \times \; -->m_i\mathbf{V}=\mathbf{R}\times <!--\; \times \; -->\underset{i}{\sum <!--\; \sum \; -->}{m}_i\mathbf{V}=\mathbf{R}\times <!--\; \times \; -->M\mathbf{V},}$

and total angular momentum for the collection of particles is finally,

The first term is the angular momentum of the center of mass relative to the origin. Similar to Single particle, below, it is the angular momentum of one particle of mass M at the center of mass moving with velocity V. The second term is the angular momentum of the particles moving relative to the center of mass, similar to Fixed center of mass, below. The result is general—the motion of the particles is not restricted to rotation or revolution about the origin or center of mass. The particles need not be individual masses, but can be elements of a continuous distribution, such as a solid body.

Rearranging equation (2) by vector identities, multiplying both terms by "one", and grouping appropriately,

gives the total angular momentum of the system of particles in terms of moment of inertia ${\displaystyle I}$ and angular velocity ${\displaystyle \mathit{\omega <!--\; \omega \; -->}}$,

Single particle caseedit

In the case of a single particle moving about the arbitrary origin,

${\displaystyle \underset{i}{\sum <!--\; \sum \; -->}{\mathbf{r}}_i\times <!--\; \times \; -->m_i{\mathbf{v}}_i=\mathbf{0},}$

${\displaystyle \underset{i}{\sum <!--\; \sum \; -->}{I}_i{\mathit{\omega <!--\; \omega \; -->}}_i=\mathbf{0},}$ and equations (2) and (3) for total angular momentum reduce to,

${\displaystyle \mathbf{L}=\mathbf{R}\times <!--\; \times \; -->m\mathbf{V}=I_R{\mathit{\omega <!--\; \omega \; -->}}_R.}$

Case of a fixed center of massedit

For the case of the center of mass fixed in space with respect to the origin,

${\displaystyle \mathbf{V}=\mathbf{0},}$

${\displaystyle \mathbf{R}\times <!--\; \times \; -->M\mathbf{V}=\mathbf{0},}$

${\displaystyle I_R{\mathit{\omega <!--\; \omega \; -->}}_R=\mathbf{0},}$ and equations (2) and (3) for total angular momentum reduce to,

${\displaystyle \mathbf{L}=\underset{i}{\sum <!--\; \sum \; -->}{\mathbf{r}}_i\times <!--\; \times \; -->m_i{\mathbf{v}}_i=\underset{i}{\sum <!--\; \sum \; -->}{I}_i{\mathit{\omega <!--\; \omega \; -->}}_i.}$