# Dynamic Modeling of Motorized Spindle System with Unbalanced Mass and Spindle Inclination
## 3 Results and Discussion
### Dynamic Characteristics of Motorized Spindle without Fault  
From Section 2.1, the general equation from the first node to the last node can be written as  
\[\mathbf{Z}_{n+1}=\mathbf{T}_{n}\mathbf{Z}_{n}=\mathbf{T}_{n}\mathbf{T}_{n-1} \mathbf{Z}_{n-1}=\cdots=\mathbf{T}_{n}\mathbf{T}_{n-1}\cdots\mathbf{T}_{1} \mathbf{Z}_{1}=\mathbf{T}_{all}\mathbf{Z}_{1} \tag{18}\]  
where \(\mathbf{Z}_{n+1}\) is the last subsegment state vector of the motorized spindle, \(\mathbf{Z}_{1}\) is the initial subsegment state vector of the motorized spindle, and \(\mathbf{T}_{all}\) is the total transfer matrix obtained by sequentially multiplying the transfer matrices of all units from the first to the last. Because both ends of the motorized spindle are free ends, we know that some parameters of the end, that is, the bending moment and shear force of these two sections, are zero. Therefore,  
\[\Delta\omega=\begin{vmatrix}a_{31}&a_{32}\\ a_{41}&a_{42}\end{vmatrix}=\Big{(}m\omega^{2}-k_{r}\Big{)}\bigg{(}k_{\theta}- \bigg{(}J_{d}-J_{p}\frac{\Omega}{\omega}\bigg{)}\omega^{2}\bigg{)}=0 \tag{19}\]  
where the critical angular velocity can be obtained by Ref. [26], \(a_{31}\), \(a_{32}\), \(a_{41}\), and \(a_{42}\) are the elements in the first or second columns of the third row and the first or second columns of the fourth row of the total transfer matrix, respectively, which contain the parameter angular velocity of \(\omega\).  
\begin{table}
\begin{tabular}{c c} \hline \hline
**Parameter** & **Value** \\ \hline Spindle length/mm & 290 \\ Mass/kg & 9.87 \\ Current/A & 10 \\ Inner diameter of motor rotor/mm & 42 \\ Outer diameter of rotating spindle/mm & 40 \\ Outer diameter of motor rotor/mm & 100 \\ Type of bearing & 71,908 CE/HCP4A \\ Maximum speed/rpm & 58,000 \\ \hline \hline \end{tabular}
\end{table}
Table 1: Parameters of high-speed motorized spindle.  
Figure 5 shows the curve of angular velocity with change and the iterative method is used to solve the problem in a small range of approximately 0. When \(\Delta\omega=0\), the value corresponding to the horizontal coordinate is the critical rotational speed. Within the specified operating speed range of the motorized spindle, there are four points satisfied as \(\Delta\omega\approx 0\). This means that the system has four critical speeds, which are 16,255.16, 24,526.62, 36,430.34, and 53,907.12 rpm.  
After the critical rotational speed is obtained, the relationship between the deflection of the initial end \(y_{1}\) and the rotation angle of the initial end \(\theta_{1}\) can be obtained by bringing the corresponding value of \(\omega\) into Equation (18). For convenience of calculation, a dimensionless calculation is usually carried out, and the deflection of the initial end \(y_{1}\) is set to 1. The angle \(\theta_{1}\) can be obtained as follows:  
\[\begin{pmatrix}y\\ \theta\end{pmatrix}^{L}=\begin{pmatrix}a_{31}&a_{32}\\ a_{41}&a_{42}\end{pmatrix}\begin{pmatrix}y\\ \theta\end{pmatrix}^{R}\Rightarrow\theta^{L}=a_{41}y^{R}+a_{42}\theta^{R}=0 \Rightarrow\theta^{R}=-\frac{a_{41}}{a_{42}}y^{R}=\alpha y^{R}\\ \theta_{1}=-\frac{a_{41}}{a_{42}}y_{1}=\alpha y_{1} \tag{20}\]  
From this, the normalized state vector of the initial end at a certain critical speed can be obtained as follows:  
\[\mathbf{Z}_{1}=\begin{bmatrix}1&\alpha&0&0\end{bmatrix}^{T} \tag{21}\]  
Equation (21) is multiplied by the element transfer matrix of each element in turn to obtain the cross-sectional section state vector of the corresponding node. At the different critical speeds, the deflection of each node can be calculated, and the mode shapes are shown in Figure 6.  
Figure 6: _Cont._  
From (a-d) in Figure 6, it can be seen that under different critical speeds, the radial displacement of the spindle changes with the axial direction of the motorized spindle, and the vibration of the motorized spindle is different at different critical speeds.