Analogous to torque, the angular momentum of a body is defined as the cross product of the position vector $\vec{r}$ and the linear momentum $\vec{p}$:
$$
\vec{L} = \vec{r} \times \vec{p}
$$
Replacing the momentum with its definition $\vec{p} = m\vec{v}$ yields:
$$
\vec{L} = m\vec{r} \times \vec{v}
$$
We have already seen that in the case of rotational motion, $\vec{v}$ is always perpendicular to $\vec{r}$. For the rotation of a rigid body about its center of mass, it is therefore sufficient to consider only the magnitude of $\vec{L}$. Integrating over infinitesimally small mass elements, as previously shown, gives:
$$
L = \int rv\,\mathrm{d}m = \int r^2\omega\,\mathrm{d}m
$$
Substituting the definition of the moment of inertia yields the relationship between angular momentum and angular velocity for a rotating rigid body:
$$
L = I\omega
$$
It should be noted that angular momentum in a closed system is conserved, just like energy and linear momentum. This can be shown by taking the time derivative of the angular momentum:
$$
\frac{\mathrm{d}\vec{L}}{\mathrm{d}t} = \frac{\mathrm{d}(\vec{r} \times \vec{p})}{\mathrm{d}t}
$$
It can be verified that the derivative of a cross product follows the same product rule as the derivative of a product of two scalars:
$$
\frac{\mathrm{d}\vec{L}}{\mathrm{d}t} = \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \times \vec{p} + \vec{r} \times \frac{\mathrm{d}\vec{p}}{\mathrm{d}t}
$$
Since the velocity $\vec{v} = \mathrm{d}\vec{r}/\mathrm{d}t$ has the same direction as the momentum $\vec{p}$, the first term vanishes, leaving:
$$
\frac{\mathrm{d}\vec{L}}{\mathrm{d}t} = \vec{r} \times \frac{\mathrm{d}\vec{p}}{\mathrm{d}t} = \vec{r} \times \vec{F}
$$
which is exactly the definition of torque. This leads to the relationship between torque and angular momentum:
$$
\vec{D} = \frac{\mathrm{d}\vec{L}}{\mathrm{d}t}
$$
To change the angular momentum of a rotating body, a torque must therefore be applied. A central force, on the other hand, leaves the angular momentum unchanged over time. From the conservation of angular momentum in a system without external forces, the following statement also follows: if the moment of inertia of a body changes, the angular velocity must increase or decrease accordingly so that $L$ remains constant. This effect is used, for example, by figure skaters who pull in their arms during a spin to increase their rotation speed.
Taking the explicit time derivative of angular momentum gives a relationship analogous to Newton's second law for translational motion:
$$
\vec{D} = I\frac{\mathrm{d}\vec{\omega}}{\mathrm{d}t} = I\vec{\alpha}
$$
Here, $\vec{\alpha}$ denotes the angular acceleration vector, which is defined as the time derivative of the angular velocity. From this, similar to the position-time law, an angular position-time law for constant angular acceleration can be derived:
$$
\varphi(t) = \frac{1}{2}\alpha t^2 + \omega t + \varphi_0
$$
Here, angular acceleration is assumed to be a scalar quantity, i.e., the direction of the rotation axis remains constant at all times.