# Dynamics Analysis of Rotating Cantilever Beams with Free End Mass
## 2 Equations of Motions  
As shown in Figure 1, the object in this study is the rotating cantilever beam system with a free end mass. A rigid center body, which rotates around the fixed axis \(O\) with angular speed \(\Omega\), has the radius \(a\). There is a rotating cantilever beam mounted on the edge of the rigid body, with the density \(\rho\), the elastic modulus \(E\), the length \(L\), the cross-sectional area \(A\), and the section moment of inertia \(I\). A lumped mass \(m\) is attached at the beam's free end. Before the cantilever beam is deformed, the neutral axis extension line passes through the rotation axis \(O\). The inertial coordinate system \(O\)-\(XY\) is established through the fixed axis \(O\) of the rigid center body, and the floating coordinate \(o\)-\(xy\) is established at the connection point \(o\). \(\tau\) is the external moment act on the center body, and \(\theta\) is the angle of the cantilever beam departure from the inertial system. Here, we treat the cantilever beam as a Euler-Bernoulli beam in order to neglect the torsional and shear deformations. This simplification makes our model more tractable. Thus, we can use the displacements at the neutral line to represent the displacements of an arbitrary point in the cantilever beam. The displacements of the arbitrary point \(P\) in the \(x\) and \(y\) direction are \(u\) and \(w\) in the floating coordinate system \(o\)-\(xy\), respectively. Then, the vector \(\mathbf{r}_{P}\) of the point \(P\) in the cantilever beam after deformation and the vector \(\mathbf{r}_{m}\) of the free end mass \(m\) are given as:  
\[\mathbf{r}_{P} =\mathbf{R}(\mathbf{r}_{a}+\mathbf{r}_{x}) \tag{1}\] \[\mathbf{r}_{m} =\mathbf{R}(\mathbf{r}_{a}+\mathbf{r}_{L}) \tag{2}\]  
where \(\mathbf{R}\) is the rotational transformation matrix, \(\mathbf{r}_{a}\) is the vector of the origin o, \(\mathbf{r}_{x}\) and \(\mathbf{r}_{L}\) are the vectors of the point \(P\) and the free end mass \(m\) in the floating coordinate system \(o\)-\(xy\), which are given as:  
\[\mathbf{R}= \begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix} ,\quad\mathbf{r}_{a}= \begin{bmatrix}a\\ 0\end{bmatrix} ,\quad\mathbf{r}_{x}= \begin{bmatrix}x+u\\ w\end{bmatrix} ,\quad\mathbf{r}_{L}= \begin{bmatrix}L+u(L)\\ w(L)\end{bmatrix}  \tag{3}\]  
The corresponding velocities of the point \(P\) and the free end mass \(m\) are given as:  
\[\dot{\mathbf{r}}_{P} =\dot{\mathbf{R}}(\mathbf{r}_{a}+\mathbf{r}_{x})+\mathbf{R}\dot{ \mathbf{r}}_{x} \tag{4}\] \[\dot{\mathbf{r}}_{m} =\dot{\mathbf{R}}(\mathbf{r}_{a}+\mathbf{r}_{L})+\mathbf{R}\dot{ \mathbf{r}}_{L} \tag{5}\]  
where a top dot means the derivation with respect to time \(t\). The details of \(\dot{\mathbf{r}}_{P}\) and \(\dot{\mathbf{r}}_{m}\) are shown as Equations (A1) and (A2) in Appendix A. Thus, the kinetic energy \(T\) is given as:  
\[T=\frac{1}{2}\int_{0}^{L}\rho A\dot{\mathbf{r}}_{P}^{T}\dot{\mathbf{r}}_{P} \mathrm{d}x+\frac{1}{2}m\dot{\mathbf{r}}_{m}^{T}\dot{\mathbf{r}}_{m} \tag{6}\]  
According to continuum mechanics, a fully normal strain-displacement relationship should be expressed as:  
Figure 1: Rotating cantilever beam system with a free end mass. and \(w_{0}\) should be fixed values in the equilibrium state. The derivatives of \(u_{0}\) and \(w_{0}\) with respect to time are zero. The perturbed deformations mean the vibration deformations of a rotating cantilever beam, are in the neighborhood of the equilibrium deformations. Hence, it is necessary to obtain the equilibrium deformations. According to the static equilibrium state, the equilibrium equations are obtained as:  
\[\int_{0}^{L}\left(EA\frac{\mathrm{d}^{2}u_{0}}{\mathrm{d}x^{2}}\right)\mathrm{ d}x=\int_{0}^{L}\left[\rho A\Omega^{2}(a+x)\right]\mathrm{d}x+m\Omega^{2}(a+L), \quad w_{0}=0 \tag{14}\]  
Here, it can be seen that the lumped mass at the free end will cause additional axial deformation due to rotation.