The circular disk
====================

The objective of this chapter is to characterize the gyroscopic matrices of an infinitely rigid circular disk, subject to a constant or variable rotation speed.

The characteristics of the disc are as follows:


* axis of the disk merged with the axis of the neutral fiber of the beam (axis :math:`\overrightarrow{x}`)


* center of gravity of the disk: :math:`C`


* inner radius: :math:`{R}_{i}`


* outer radius: :math:`{R}_{e}`


* thickness: :math:`h`


* assumed uniform density: :math:`\rho`

Deduced values:


* disk mass: :math:`M\mathrm{=}\pi \rho h({R}_{e}^{2}\mathrm{-}{R}_{i}^{2})`


* mass moment of inertia/axes :math:`y` or :math:`z` calculated at the center of gravity :math:`C`:

:math:`{I}_{\text{yz}}\mathrm{=}\frac{M}{\text{12}}(3\text{.}{R}_{e}^{2}+3\text{.}{R}_{i}^{2}+{h}^{2})`


* mass moment of inertia with respect to the :math:`x` axis calculated at the center of gravity :math:`C`:

:math:`{I}_{x}\mathrm{=}\frac{M}{2}({R}_{e}^{2}+{R}_{i}^{2})`

Notes:


* Since the axes :math:`C\overrightarrow{x}`, :math:`C\overrightarrow{y}` and :math:`C\overrightarrow{z}` are the main axes of inertia of the disk, the inertia products :math:`{I}_{\mathit{xy}}`, :math:`{I}_{\mathit{yz}}` and :math:`{I}_{\mathit{xz}}` are zero.


* The symmetry of the disk with respect to the axes :math:`C\overrightarrow{y}` and :math:`C\overrightarrow{z}` requires: :math:`{I}_{\mathit{yz}}\mathrm{=}{I}_{y}\mathrm{=}{I}_{z}`

The displacement of the center of the disk is given by: :math:`u\text{.}\overrightarrow{x}+v\text{.}\overrightarrow{y}+w\text{.}\overrightarrow{z}`

We note:


* :math:`{\overrightarrow{\Omega }}_{R/\mathrm{R0}}`: the speed vector of rotation of the disk


* :math:`\overrightarrow{x}\text{.}{\overrightarrow{\Omega }}_{R\mathrm{/}\mathit{R0}}\mathrm{=}\dot{\phi }(t)`: natural rotation speed




Calculation of the kinetic energy of the disk
------------------------

We calculate the kinetic energy of the disk by applying the Huygens formula:

:math:`T\mathrm{=}\frac{1}{2}M\text{.}{({\overrightarrow{V}}_{C,D\mathrm{/}\mathit{R0}})}^{2}+\frac{1}{2}{\overrightarrow{\Omega }}_{R\mathrm{/}\mathit{R0}}\text{.}\left[J\right]\text{.}{\overrightarrow{\Omega }}_{R\mathrm{/}\mathit{R0}}`

:math:`T\mathrm{=}\frac{1}{2}M\text{.}{({\dot{u}}^{2}+{\dot{v}}^{2}+{\dot{w}}^{2})}^{2}+\frac{1}{2}{\overrightarrow{\Omega }}_{R\mathrm{/}\mathit{R0}}\text{.}\left[J\right]\text{.}{\overrightarrow{\Omega }}_{R\mathrm{/}\mathit{R0}}`

with: :math:`\left[J\right]\mathrm{=}\left[\begin{array}{ccc}{I}_{x}& 0& 0\\ 0& {I}_{y}& 0\\ 0& 0& {I}_{z}\end{array}\right]` with :math:`{I}_{\mathit{yz}}\mathrm{=}{I}_{y}\mathrm{=}{I}_{z}`

Expanding the previous expression, we get:

:math:`T\mathrm{=}\frac{1}{2}\rho \text{.}{({\dot{u}}^{2}+{\dot{v}}^{2}+{\dot{w}}^{2})}^{2}+\frac{1}{2}{I}_{yz}\text{.}({\dot{\theta }}_{y}^{2}+{\dot{\theta }}_{z}^{2})+\frac{1}{2}{I}_{x}\text{.}({\dot{\phi }}^{2}+2{\dot{\theta }}_{y}\text{.}\dot{\phi }\text{.}{\theta }_{z})`

The various kinetic energy terms represent:


* for the first term, the kinetic energy of translation,


* for the second term, the kinetic energy of rotation,


* for the term :math:`\frac{1}{2}{I}_{x}\text{.}{\dot{\phi }}^{2}`, "clean" rotational energy,


* and for the term :math:`{I}_{x}\text{.}({\dot{\theta }}_{y}\text{.}\dot{\phi }\text{.}{\theta }_{z})`, the gyroscopic effect.




Calculation of equilibrium equations
--------------------------------

The Lagrange equations are used to formulate the dynamic equilibrium of the disk. In this specific case, the deformation energy is zero (infinitely rigid disk) and no external force is considered, so we have:

:math:`\frac{d}{\text{dt}}(\frac{\mathrm{\partial }T}{\mathrm{\partial }{\dot{q}}_{i}})\mathrm{-}\frac{\mathrm{\partial }T}{\mathrm{\partial }{q}_{i}}\mathrm{=}0` with :math:`\mathrm{\langle }q\mathrm{\rangle }\mathrm{=}\mathrm{\langle }uvw{\theta }_{y}{\theta }_{z}\mathrm{\rangle }`: vector of the degrees of freedom of the disk element.

We do not take into account the degree of freedom :math:`\phi` because we consider that the natural speed of rotation is imposed and therefore known. The following equations are then obtained:

 :math:`\mathrm{\{}\begin{array}{c}\frac{d}{\text{dt}}(\frac{\mathrm{\partial }T}{\mathrm{\partial }\dot{u}})\mathrm{-}\frac{\mathrm{\partial }T}{\mathit{\delta u}}\mathrm{=}M\text{.}\ddot{u}\\ \frac{d}{\text{dt}}(\frac{\mathrm{\partial }T}{\mathrm{\partial }\dot{v}})\mathrm{-}\frac{\mathrm{\partial }T}{\mathit{\delta v}}\mathrm{=}M\text{.}\ddot{v}\\ \frac{d}{\text{dt}}(\frac{\mathrm{\partial }T}{\mathrm{\partial }\dot{w}})\mathrm{-}\frac{\mathrm{\partial }T}{\mathit{\delta w}}\mathrm{=}M\text{.}\ddot{w}\\ \frac{d}{\text{dt}}(\frac{\mathrm{\partial }T}{\mathrm{\partial }{\dot{\theta }}_{y}})\mathrm{-}\frac{\mathrm{\partial }T}{\mathrm{\partial }{\theta }_{y}}\mathrm{=}{I}_{\text{yz}}\text{.}{\ddot{\theta }}_{y}+{I}_{x}\text{.}\dot{\phi }\text{.}{\dot{\theta }}_{z}+{I}_{x}\text{.}\ddot{\phi }\text{.}{\theta }_{z}\\ \frac{d}{\text{dt}}(\frac{\mathrm{\partial }T}{\mathrm{\partial }{\dot{\theta }}_{z}})\mathrm{-}\frac{\mathrm{\partial }T}{\mathrm{\partial }{\theta }_{z}}\mathrm{=}{I}_{\text{yz}}\text{.}{\ddot{\theta }}_{z}\mathrm{-}{I}_{x}\text{.}\dot{\phi }\text{.}{\dot{\theta }}_{y}\end{array}`

These equations can be put in the form:

:math:`\left[M\right]\mathrm{\langle }\ddot{q}\mathrm{\rangle }+\left[{C}_{\mathit{gyro}}\right]\mathrm{\langle }\dot{q}\mathrm{\rangle }+(\left[K\right]+\left[{K}_{\mathit{gyro}}\right])\mathrm{\langle }q\mathrm{\rangle }\mathrm{=}\mathrm{\langle }0\mathrm{\rangle }`

The gyroscopic damping matrix of the disk is obtained from moment of inertia :math:`{I}_{x}`. It is antisymmetric, and its contribution must be multiplied by the natural angular speed :math:`\dot{\varphi }`.

:math:`\left[{C}_{\mathit{gyro}}\right]=\dot{\varphi }\text{.}\left[\begin{array}{cccccc}0& 0& 0& -& 0& 0\\ 0& 0& 0& -& 0& 0\\ 0& 0& 0& -& 0& 0\\ -& -& -& -& -& -\\ 0& 0& 0& -& 0& {I}_{\text{x}}\\ 0& 0& 0& -& -{I}_{\text{x}}& 0\end{array}\right]`

with :math:`\langle uvw{\theta }_{x}{\theta }_{y}{\theta }_{z}\rangle` vector of the degrees of freedom of the disk element and such that: :math:`{\dot{\theta }}_{x}\mathrm{=}\dot{\phi }`

The dash corresponds to the degree of freedom of rotation along the axis of the beam and obviously leads to zero terms.

The gyroscopic stiffness matrix of the disk is also obtained from moment of inertia :math:`{I}_{x}`. Its contribution must be multiplied by the natural angular acceleration :math:`\ddot{\phi }`.

:math:`\left[{K}_{\mathit{gyro}}\right]\mathrm{=}\ddot{\phi }\text{.}\left[\begin{array}{cccccc}0& 0& 0& \mathrm{-}& 0& 0\\ 0& 0& 0& \mathrm{-}& 0& 0\\ 0& 0& 0& \mathrm{-}& 0& 0\\ \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}\\ 0& 0& 0& \mathrm{-}& 0& {I}_{\text{x}}\\ 0& 0& 0& \mathrm{-}& 0& 0\end{array}\right]`