The angular momentum of an extended system depends on the point about which it is calculated. A useful decomposition separates the motion of the center of mass from motion relative to the center of mass.

Let $O$ be a chosen origin. For particle $i$, write

$$\mathbf r_i=\mathbf R_{\rm CM}+\mathbf r_i',$$

and

$$\mathbf v_i=\mathbf V_{\rm CM}+\mathbf v_i',$$

where primed quantities are measured relative to the center of mass.

The total angular momentum about $O$ is

$$\mathbf L_O=\sum_i\mathbf r_i\times m_i\mathbf v_i.$$

Substituting the decompositions gives four kinds of terms:

$$\mathbf L_O =\sum_i(\mathbf R_{\rm CM}+\mathbf r_i') \times m_i(\mathbf V_{\rm CM}+\mathbf v_i').$$

Using

$$\sum_i m_i=M,$$

$$\sum_i m_i\mathbf r_i'=\mathbf0,$$

and

$$\sum_i m_i\mathbf v_i'=\mathbf0,$$

the cross terms vanish. Therefore

$$\boxed{\mathbf L_O =\mathbf R_{\rm CM}\times M\mathbf V_{\rm CM} +\mathbf L_{\rm CM}},$$

where

$$\boxed{\mathbf L_{\rm CM}=\sum_i\mathbf r_i'\times m_i\mathbf v_i'}$$

is the angular momentum measured about the center of mass.

More explicitly, if $\mathbf R_{\rm CM/O}$ is the vector from $O$ to the center of mass and $\mathbf P=M\mathbf V_{\rm CM}$ is total linear momentum,

$$\boxed{\mathbf L_O =\mathbf R_{\rm CM/O}\times\mathbf P+\mathbf L_{\rm CM}}.$$

Orbital and internal parts

The term

$$\mathbf R_{\rm CM/O}\times\mathbf P$$

is the angular momentum associated with motion of the center of mass about the chosen origin. It is sometimes called the orbital part.

The term $\mathbf L_{\rm CM}$ describes motion relative to the center of mass. For a rigid body rotating in a plane about a principal axis through its center of mass,

$$\boxed{\mathbf L_{\rm CM}=I_{\rm CM}\boldsymbol\omega}.$$

The total angular momentum can therefore contain both translation of the body as a whole and rotation about its center.

Worked example: translating and spinning disk

A uniform disk of mass

$$M=2.0,\mathrm{kg}$$

and radius

$$R=0.50,\mathrm m$$

has center-of-mass velocity

$$\mathbf V_{\rm CM}=3.0\hat{\mathbf y},\mathrm{m/s}$$

at an instant when its center is located at

$$\mathbf R_{\rm CM/O}=4.0\hat{\mathbf x},\mathrm m.$$

It also rotates counterclockwise with

$$\boldsymbol\omega=5.0\hat{\mathbf z},\mathrm{rad/s}.$$

The total momentum is

$$\mathbf P=M\mathbf V_{\rm CM}=6.0\hat{\mathbf y},\mathrm{kg,m/s}.$$

The orbital contribution about $O$ is

$$\mathbf R_{\rm CM/O}\times\mathbf P =(4.0\hat{\mathbf x})\times(6.0\hat{\mathbf y}) =24\hat{\mathbf z},\mathrm{kg,m^2/s}.$$

For a uniform disk,

$$I_{\rm CM}=\frac12MR^2 =\frac12(2.0)(0.50)^2 =0.25,\mathrm{kg,m^2}.$$

Thus the internal rotational contribution is

$$\mathbf L_{\rm CM} =I_{\rm CM}\boldsymbol\omega =(0.25)(5.0)\hat{\mathbf z} =1.25\hat{\mathbf z},\mathrm{kg,m^2/s}.$$

The total angular momentum about $O$ is therefore

$$\boxed{\mathbf L_O =25.25\hat{\mathbf z},\mathrm{kg,m^2/s}}.$$

Most of the angular momentum in this example comes from the translation of the center of mass about $O$, not from the disk's spin.

Choice of origin matters

If the origin is moved to the center of mass, then

$$\mathbf R_{\rm CM/CM}=\mathbf0,$$

so only the internal angular momentum remains.

If the total momentum is zero,

$$\mathbf P=\mathbf0,$$

then

$$\mathbf L_O=\mathbf L_{\rm CM}$$

for any origin related by a simple translation. Thus angular momentum becomes independent of the translated origin when total linear momentum vanishes.

The decomposition is the angular-momentum counterpart of separating kinetic energy into translation of the center of mass plus motion relative to it. It is useful whenever a system both translates and rotates, including rolling bodies, orbiting-and-spinning systems, and collision problems.