PC1672 Advanced dynamics

4.1 Angular momentum and angular velocity

The important dynamical quantities that we use to describe rotational motion are all vectors. The directions of these vectors do not point along the motions of particles but rather along axes of rotation or twisting. The motion of a spinning system like a gyroscope, which can look surprising at first, can be best understood by thinking about the directions of these rotational vectors.

The angular momentum of a single particle is defined the moment of its linear momentum about our chosen origin:

$\begin{displaymath}{\bf L}={\bf x}\times{\bf p}\end{displaymath}$

where ${\bf x}$ is its position with respect to the origin and ${\bf p}$ is its momentum.

The torque acting on the particle is defined similarly as the moment of the force ${\bf F}$ on the particle:

$\begin{displaymath}\hbox{\boldmath {$\tau$}}={\bf x}\times{\bf F}\end{displaymath}$

The angular equation of motion is

$\begin{displaymath}\dot{\bf L}=\hbox{\boldmath {$\tau$}}\end{displaymath}$

(torque is rate of change of angular momentum).

We shall be interested in rigid rotations, where every particle lies at a fixed distance from the origin and at a fixed angle to the axis of rotation. For such rotations we define a vector angular velocity which points along the axis of rotation. The (linear) velocity of a particle rotating in this way is

$\begin{displaymath}{\bf v}=\hbox{\boldmath {$\omega$}}\times{\bf x}\end{displaymath}$

The angular momentum of our rigidly rotating particle is thus

$\begin{displaymath}{\bf L}=m{\bf x}\times(\hbox{\boldmath {$\omega$}}\times{\bf x})\end{displaymath}$

The vector triple product can be evaluated to give

$\begin{displaymath}{\bf L}=m\left[{\bf x}^2\hbox{\boldmath {$\omega$}}-({\bf x}\cdot \hbox{\boldmath {$\omega$}}){\bf x}\right]\end{displaymath}$

This shows that, in general, angular momentum and angular velocity point along different directions.

Since ${\bf L}$ lies in the same plane as ${\bf x}$ , it rotates as ${\bf x}$ rotates. Hence the system must be subject to a torque

$\begin{displaymath}\hbox{\boldmath {$\tau$}}=\hbox{\boldmath {$\omega$}}\times{\bf L}\end{displaymath}$

However, if we add another equal mass the same distance away but on the opposite side of the axis, then the total angular momentum will be parallel to the angular velocity. In such a case we say that the system is dynamically balanced; no torque is needed to maintain its rotation.

Textbook references

K&K 7.2
F&C (6.4, 7.2)
B&O 5.1, 6.4
M 1.15, 2.5, 12.3 (1.11)
M&T 1.15, 2.5, 11.3 (1.11)
B 6.2, 6.3 for vector products

Mike Birse
17th May 2000