Angular Momentum of a Particle and of a System of Particles

The angular momentum of a single particle about a chosen reference point $O$ is defined, using the

same vector (cross) product introduced in Section 5.5, as

where $\vec r$ is the particle's position vector measured from $O$ and $\vec p = m\vec v$ is its linear

momentum. Just as torque is the rotational analogue of force, angular momentum is the rotational analogue

of linear momentum, and its magnitude is $l = rp\sin\theta = mvr\sin\theta$, with $\theta$ the angle

between $\vec r$ and $\vec p$ (equivalently, between $\vec r$ and $\vec v$, since $\vec p$ is simply a

positive scalar multiple of $\vec v$).

For the particular case of a particle moving in a circle of radius $r$ about the very point $O$ chosen as

the reference (see the accompanying figure), the velocity -- and hence the momentum -- is always exactly

tangent to the circle, so $\vec r$ and $\vec p$ are always perpendicular ($\theta = 90^\circ$, $\sin\theta = 1$), giving the simpler and very frequently used result

the second equality following from $v = \omega r$ and the definition of moment of inertia introduced later

in Section 5.10 (for a single particle at distance $r$ from the axis, $I = mr^2$).

Total angular momentum of a system of particles. Exactly as with linear momentum (Section 5.4), the

total angular momentum of a whole system of particles about a chosen point $O$ is simply the vector sum of

the individual angular momenta of every particle in the system, $\vec L = \sum_i \vec l_i$.

Newton's second law in rotational form. Differentiating $\vec l = \vec r\times\vec p$ with respect to

time and using $d\vec p/dt = \vec F$ (Newton's second law for the particle) gives, after the term

involving $d\vec r/dt \times \vec p$ vanishes (since $d\vec r/dt = \vec v$ is always parallel to $\vec p = m\vec v$, and the cross product of two parallel vectors is zero),

$\frac{d\vec l}{dt} = \vec r \times \vec F = \vec\tau$ …