For a particle, angular momentum about an origin is

$$\mathbf L=\mathbf r\times\mathbf p.$$

When a system rotates about a fixed axis, the component of angular momentum along that axis has a particularly simple form.

Let the rotation axis be the $z$-axis. A mass element $m_i$ at perpendicular distance $r_{\perp i}$ from the axis moves with tangential speed

$$v_i=r_{\perp i}|\omega|.$$

Its angular-momentum component along the axis is

$$L_{z,i}=m_i r_{\perp i}^2\omega$$

with the sign set by the rotation direction. Summing over the body,

$$L_z=\left(\sum_i m_i r_{\perp i}^2\right)\omega.$$

Since

$$I_z=\sum_i m_i r_{\perp i}^2,$$

we obtain

$$\boxed{L_z=I_z\omega}.$$

For a continuous body, the same result follows from

$$I_z=\int r_\perp^2,dm.$$

Component relation versus vector relation

The equation

$$L_z=I_z\omega$$

always describes the angular-momentum component along the fixed rotation axis. The stronger vector statement

$$\mathbf L=I\boldsymbol\omega$$

is valid when the rotation is about a principal axis, such as a symmetry axis of a disk or cylinder. For a general three-dimensional rotation, $\mathbf L$ and $\boldsymbol\omega$ need not be parallel.

Torque and the familiar rotational equation

The angular-momentum law gives

$$\tau_{z,\rm ext}=\frac{dL_z}{dt}.$$

If the moment of inertia about the fixed axis is constant,

$$\tau_{z,\rm ext}=I_z\frac{d\omega}{dt} =I_z\alpha.$$

Thus the familiar fixed-axis equation

$$\tau=I\alpha$$

is a special case of the more general angular-momentum relation.

Conservation of angular momentum

If the net external torque about the axis is zero,

$$\tau_{z,\rm ext}=0,$$

then

$$\boxed{L_z=\text{constant}}.$$

If the system changes its mass distribution while remaining isolated about that axis, the moment of inertia can change even though angular momentum does not. Then

$$\boxed{I_i\omega_i=I_f\omega_f}.$$

Internal forces can rearrange the mass distribution but cannot change the system's total angular momentum when they produce no net external torque.

Worked example

A rotating system initially has

$$I_i=4.0,\mathrm{kg,m^2}$$

and

$$\omega_i=2.0,\mathrm{rad/s}.$$

Its angular momentum is

$$L=I_i\omega_i=(4.0)(2.0)=8.0,\mathrm{kg,m^2/s}.$$

Suppose internal motion reduces the moment of inertia to

$$I_f=2.0,\mathrm{kg,m^2}$$

while external torque about the axis is negligible. Conservation gives

$$8.0=(2.0)\omega_f,$$

so

$$\boxed{\omega_f=4.0,\mathrm{rad/s}}.$$

The angular speed doubles because the same angular momentum is carried by a smaller rotational inertia.

Fixed-axis angular momentum therefore connects particle angular momentum, moment of inertia, rotational dynamics, and conservation in a single reusable relation.