## 3 3D Dynamics Deduction
### Dynamic Equilibrium Equations
#### 3.3.2 Moment Balance around the \(\mathbf{x}\)-Axis of the \(<xyz>_{C}\) Coordinate System  
The \(<xyz>_{O}\) and the \(<xyz>_{C}\) are two parallel coordinate systems. In the coordinate system \(<xyz>_{C}\), the angular velocity \(\mathbf{\Omega}_{C}\) around the \(\mathbf{z}\)-axis, the centripetal forces \(\mathbf{F}_{m}\) and \(\mathbf{F}_{Bi}(i=1\sim n)\) of the rotor mass center and the balls are the same as those in \(<xyz>_{O}\), respectively. \(\mathbf{\Omega}_{C}=\mathbf{\Omega}_{O}= \begin{bmatrix}0&0&\omega\end{bmatrix} ^{T}\), \(\mathbf{F}_{m}=\mathbf{F}_{Om}\) (Equation (8)), and \(\mathbf{F}_{Bi}=\mathbf{F}_{OBi}\) (Equation (9)). The equilibrium position vector of the mass center of the rotor and the \(i\)-th ball in \(<xyz>_{C}\) is \(\hat{\mathbf{r}}_{m}\) and \(\hat{\mathbf{r}}_{Bi}(i=1\sim n)\). To inverse transformation of \(\hat{\mathbf{r}}_{m}^{\prime\prime}\) and \(\hat{\mathbf{r}}_{Bi}^{\prime\prime}\) (Equations (3) and (5)) or to translation of \(\hat{\mathbf{r}}_{Om}\) and \(\hat{\mathbf{r}}_{OBi}\) (Equations (6) and (7)) can be obtained as \[\hat{\mathbf{r}}_{m}=\mathbf{T}_{1}^{-1}\mathbf{T}_{2}^{-1}\hat{\mathbf{r}}_{m}^{\prime\prime}=\hat{ \mathbf{r}}_{Om}-\hat{\mathbf{r}}_{OC}= \begin{bmatrix}\epsilon\cos\psi\cos\hat{\gamma}_{2 }\\ \epsilon\sin\psi\cos\hat{\gamma}_{1}+\epsilon\cos\psi\sin\hat{\gamma}_{1}\sin \hat{\gamma}_{2}\\ \epsilon\sin\psi\sin\hat{\gamma}_{1}-\epsilon\cos\psi\cos\hat{\gamma}_{1}\sin \hat{\gamma}_{2}\end{bmatrix} \text{ and } \tag{15}\]  
\[\hat{\mathbf{r}}_{Bi}=\mathbf{T}_{1}^{-1}\mathbf{T}_{2}^{-1}\hat{\mathbf{r}}_{Bi}^{\prime\prime }=\hat{\mathbf{r}}_{OBi}-\hat{\mathbf{r}}_{OC}= \begin{bmatrix}R\cos\hat{\beta}_{i} \cos(\hat{\gamma}_{2}+\theta)\\ R\sin\hat{\beta}_{i}\cos\hat{\gamma}_{1}+R\cos\hat{\beta}_{i}\sin\hat{\gamma}_ {1}\sin(\hat{\gamma}_{2}+\theta)\\ R\sin\hat{\beta}_{i}\sin\hat{\gamma}_{1}-R\cos\hat{\beta}_{i}\cos\hat{\gamma}_ {1}\sin(\hat{\gamma}_{2}+\theta)\end{bmatrix} ,i=1\sim n. \tag{16}\]  
The absolute angular velocity \(\mathbf{\Omega}\) of the principal axis of inertia of the eccentric rotor in \(<x^{\alpha}y^{\alpha}z^{\alpha}>_{C}\) can transform by \(\mathbf{\Omega}_{C}\), and rotor angular momentum \(\hat{\mathbf{H}}\) can be obtained from \(\hat{\mathbf{\Omega}}\) and the principal moment of inertia tensor \(\mathbf{J}\). The torque due to the angular momentum of the eccentric rotor is expressed in the cross-product term \(\hat{\mathbf{\Omega}}\times\hat{\mathbf{H}}\), where  
\[\hat{\mathbf{\Omega}}=\mathbf{T}_{2}\mathbf{T}_{1}\mathbf{\Omega}_{C}=\mathbf{T}_{2}\mathbf{T}_{1} \begin{bmatrix}0\\ 0\\ \omega\end{bmatrix} = \begin{bmatrix}-\omega\text{cos}\,\hat{\gamma}_{1}\text{sin }\,\hat{\gamma}_{2}\\ \omega\text{sin}\,\hat{\gamma}_{1}\\ \omega\text{cos}\,\hat{\gamma}_{1}\text{cos}\,\hat{\gamma}_{2}\end{bmatrix}  \text{ and } \tag{17}\]  
\[\hat{\mathbf{H}}=\mathbf{J}\hat{\mathbf{\Omega}}= \begin{bmatrix}J_{T}&0&0\\ 0&J_{T}&0\\ 0&0&J_{P}\end{bmatrix} \hat{\mathbf{\Omega}}. \tag{18}\]  
The support reaction moment of the torsion spring to the \(\mathbf{x}\)-axis in dynamic balance is \(k_{T}\hat{\gamma}_{1}\). Therefore, the 3D spatial moment balanced on the \(\mathbf{x}\)-axis in the coordinate system \(<\mathbf{xyz}>_{C}\) is expressed as  
\[ \begin{bmatrix}\hat{\mathbf{r}}_{m}\times F_{m}+\sum_{i=1}^{n}\hat{\mathbf{r}}_{Bi} \times F_{Bi}+\mathbf{T}_{1}^{-1}\mathbf{T}_{2}^{-1}\big{(}\hat{\mathbf{\Omega}}\times \hat{\mathbf{H}}\big{)}\end{bmatrix} . \begin{bmatrix}1&0&0\end{bmatrix} ^{T}+k_{T} \hat{\gamma}_{1}=0. \tag{19}\]  
The moment equilibrium equation of the system for the \(\mathbf{x}\)-axis in \(<\mathbf{xyz}>_{C}\) can be listed as  
\[