[NOTES/CM-09011] General Displacement of a Rigid Body

If two points of the rigid body are held fixed, the rigid body can rotate about an axis passing through the two points. The direction of angular velocity remains constant all the time only its magnitude may vary. The angular momentum $\vec{L}$ will be parallel to the angular velocity $\vec{\Omega}$
\begin{equation} \vec{L}=I\vec{\Omega} \end{equation}
where $I$ is the moment of inertia about the axis of rotation. In this case the EOM is
\begin{equation} \frac{d\vec{L}}{dt}=\vec{T}, \end{equation} or, \begin{equation} I\frac{d\vec{\Omega}}{dt}=\vec{T}. \end{equation}
In fact, $\vec{L},\vec{\Omega}$ and $\vec{T}$ are all parallel and it is hardly necessary to use vectors. One may simply write $I\frac{d{\Omega}}{dt}={T}$, where $T$ is the comp of torque along the axis of rotation. An example of this type of motion is rotation of bicycle wheel about its axle, if the end points of the axle are held fixed. %
If only one point of a rigid body is hold fixed, the general motion of the body is a rotation about some axis passing through the selected point. This axis is described by a unit vector $\hat{n}$, requiring two independent parameters. The rotation requires an additional coordinate viz, the angle of rotation. Thus this case one needs three generalised coordinates. An example of motion of the type is motion of a top on a plane such that the point of contacts of the top and of the plane do not move.

If a rigid is body is free to move in a force field, for example a space station or a satellite, the most general motion of the body consists, in rotation about the center of mass and translation of the center of mass. Thus six generalised coordinates are needed to describe it. The total momentum, kinetic energy and angular momentum can be written as sums of corresponding quantities for center of mass ($\vec R$) and that for relative coordinates \(R_\alpha\). For a multi particle system we have
Total momentum $=\vec{P}=\sum_\alpha M_\alpha\dot{\vec{r_\alpha}}$$=M\dot{\vec{R}}$
Total Energy $=\frac{1}{2}M\dot{\vec{R^2}}+\sum_{\alpha}\frac{1}{2} M_{\alpha}\dot{\vec{R_\alpha}^{2}}$
Total angular Momentum $=M\bar{R}\times \bar{V}+\sum_{\alpha}m_{\alpha}\vec{R_\alpha}\times\dot{\vec{R}}_\alpha$

The summations $\sum_\alpha$ refer to the intrinsic or relative motion. These results are presented in a tabular form below.