Historical Prospective Objective, scope and limitations

Rigid body angular momentum

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Angular momentum and torque

Rigid body angular momentum

The most general equation for rotation of a rigid body in three dimensions about an arbitrary origin O with axes x, y, z is

$m b_{g/o} \times \frac{\mathrm{d}^2 r_o}{\mathrm{d}t^2} + \frac{\mathrm{d}(\mathbf{i}\boldsymbol{\omega})}{\mathrm{d}t} = \sum_{j=1}^n \tau_{o,j}$

where the moment of inertia tensor, $\mathbf{i}$ , is given by

$\mathbf{i} = \begin{pmatrix} i_{xx} & i_{xy} & i_{xz} \\ i_{yx} & i_{yy} & i_{yz} \\ i_{zx} & i_{zy} & i_{zz} \end{pmatrix}$

$\mathbf{i} = \begin{pmatrix} \int (y^2+z^2)\, \mathrm{d}m & -\int xy\, \mathrm{d}m & -\int xz\, \mathrm{d}m\\ -\int xy\, \mathrm{d}m & \int (x^2+z^2)\, \mathrm{d}m & -\int yz\, \mathrm{d}m \\ -\int xz\, \mathrm{d}m & -\int yz\, \mathrm{d}m & \int (x^2+y^2)\, \mathrm{d}m \end{pmatrix}$

Given that Euler's rotation theorem states that there is always an instantaneous axis of rotation, the angular velocity, $\boldsymbol{\omega}$ , can be given by a vector over this axis

$\quad \boldsymbol{\omega} = \omega_x \mathbf{\hat{i}} + \omega_y \mathbf{\hat{j}} + \omega_z \mathbf{\hat{k}}$

where $\scriptstyle{(\mathbf{\hat{i}},\ \mathbf{\hat{j}},\ \mathbf{\hat{k}})}$ is a set of mutually perpendicular unit vectors fixed in a reference frame.

Rotating a rigid body is equivalent to rotating a Poinsot ellipsoid.

Angular momentum and torque

Similarly, the angular momentum $\mathbf{l}$ for a system of particles with linear momenta p_i and distances r_i from the rotation axis is defined

$\mathbf{l} = \sum_{i=1}^{n} \mathbf{r}_{i} \times \mathbf{p}_{i} = \sum_{i=1}^{n} m_{i} \mathbf{r}_{i} \times \mathbf{v}_{i}$

For a rigid body rotating with angular velocity ω about the rotation axis $\mathbf{\hat{n}}$ (a unit vector), the velocity vector $\mathbf{v}_{i}$ may be written as a vector cross product

$\mathbf{v}_{i} = \omega \mathbf{\hat{n}} \times \mathbf{r}_{i} \ \stackrel{\mathrm{def}}{=}\ \boldsymbol\omega \times \mathbf{r}_{i}$

Where angular velocity vector $\boldsymbol\omega \ \stackrel{\mathrm{def}}{=}\ \omega \mathbf{\hat{n}}$

$\mathbf{r}_{i}$ is the shortest vector from the rotation axis to the point mass.

Substituting the formula for $\mathbf{v}_{i}$ into the definition of $\mathbf{l}$ yields

$\mathbf{l} = \sum_{i=1}^{n} m_{i} \mathbf{r}_{i} \times (\boldsymbol\omega \times \mathbf{r}_{i}) = \boldsymbol\omega \sum_{i=1}^{n} m_{i} r_{i}^{2} = i \omega \mathbf{\hat{n}}$

Where we have introduced the special case that the position vectors of all particles are perpendicular to the rotation axis (e.g., a flywheel): $\boldsymbol\omega \cdot \mathbf{r}_{i} = 0$ .

The torque $\mathbf{n}$ is defined as the rate of change of the angular momentum $\mathbf{l}$

$\mathbf{n} \ \stackrel{\mathrm{def}}{=}\ \frac{d\mathbf{l}}{dt}$

If I is constant (because the inertia tensor is the identity, because we work in the intrinsically frame, or because the torque is driving the rotation around the same axis $\mathbf{\hat{n}}$ so that I is not changing) then we may write

$\mathbf{n} \ \stackrel{\mathrm{def}}{=}\ i \frac{d\omega}{dt}\mathbf{\hat{n}} = i \alpha \mathbf{\hat{n}}$

Where α is called the angular acceleration (or rotational acceleration) about the rotation axis $\mathbf{\hat{n}}$ .

Notice that if I is not constant in the external reference frame (i.e. the three main axes of the body are different) then we cannot take the I outside the derivate. In these cases we can havetorque-free precession.

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