Rotation about a Fixed Axis#

We will observe the rotation of a system of rigid bodies on a fixed axis (\(\dot{\mathbf R} = 0\)).

Angular Momentum#

The angular momentum of the system is only the spin angular momentum so let \(\mathbf L = \mathbf S\)

\[ \mathbf L = \sum{\mathbf r_\alpha \times m_\alpha \mathbf v_\alpha} \]

  • \(\mathbf r_\alpha\) : Position vector relative to the CM

  • \(\mathbf v_\alpha\) : Velocity vector relative to the CM such that \(\mathbf v_\alpha \equiv \dot {\mathbf r_\alpha}\)

Since the rigid bodies rotating it is best to use angular frequency in stead of velocity,

\[\begin{split} \mathbf v_\alpha = \boldsymbol{\omega}_\alpha \times \mathbf r_\alpha \\ v_\alpha = \rho_\alpha \omega\\ \rho_\alpha \equiv \sqrt{x_\alpha^2 + y_\alpha^2}\\ \mathbf{L} = \sum{m_\alpha \mathbf r_\alpha \times \left(\boldsymbol{\omega}_\alpha \times \mathbf r_\alpha\right)} \\ = \sum{m_\alpha \left[\boldsymbol{\mathbf \omega}_\alpha (\mathbf r_\alpha \cdot \mathbf r_\alpha) + \mathbf{r}_\alpha(\mathbf{r}_\alpha \cdot \boldsymbol \omega_\alpha)\right]} \\ \boxed{\mathbf{L} = \sum{m_\alpha \rho^2\boldsymbol{\mathbf \omega}_\alpha} }\\ \mathbf{L} = \sum{m_\alpha \left(-x_\alpha z_\alpha, y_\alpha z_\alpha, x_\alpha^2 + y_\alpha^2 \right)} \omega \\ \boxed{\mathbf{L} = \sum{\left(-m_\alpha x_\alpha z_\alpha, m_\alpha y_\alpha z_\alpha,I_z\right)\omega}}\\ \boxed{I_z = \sum{m_\alpha \rho_\alpha^2}} \end{split}\]

Kinetic Energy#

The kinetic energy is,comprised only of the rotational` kinetic energy so,

\[\begin{split} T = \frac{1}{2}\sum{m_\alpha v_\alpha^2} = \frac{1}{2}\sum{m_\alpha^2\rho_\alpha^2 \omega^2}\\ \boxed{T = \frac{1}{2}I_z\omega^2} \end{split}\]