The first step is to determine the rate of the angular momentum of the disk. This will give the torque it generates against the spinning shaft. Using free body diagram the reactions on the beam are found.
Let the body fixed coordinates C.S. have its origin at \(O\) and attached to the shaft. Hence C.S. rotates along with the shaft as in the following diagram
In the relative angular momentum method the equation of motion of \(m\) is found from\[ \mathbf{M}_{o}=\mathbf{\dot{h}}_{o}+m\boldsymbol{\rho }\times \mathbf{\ddot{r}}_{o}\] Where \(\mathbf{M}_{o}\) is torque around \(o\) and \(\mathbf{h}_{p}\) is the angular momentum of the disk relative to the body fixed c.s. and \(\boldsymbol{\rho }\) is the position vector from \(o\) to the center of mass of disk, and \(\mathbf{\ddot{r}}_{o}\) is the absolute acceleration vector of the reference point \(o\). But since the reference point \(o\) is fixed in space in this problem then \(\mathbf{\ddot{r}}_{o}=0\) and the above reduces to\[ \mathbf{M}_{o}=\mathbf{\dot{h}}_{o}\] Where \begin{equation} d\mathbf{h}_{o}=\boldsymbol{\rho }\times \boldsymbol{\dot{\rho }}dm \tag{1} \end{equation} \(dm\) is small unit mass of the disk given by \[ dm=\frac{m}{\pi r^{2}}dA \] where \(dA\) is the small area of the disk to be integrated over. Let \(s=\left \vert \boldsymbol{\rho }\right \vert \,\ \)be the length of the position vector from \(O\), hence\begin{align*} \boldsymbol{\rho } & =-s\sin \phi \mathbf{k}+s\cos \phi \mathbf{j}\\ \boldsymbol{\dot{\rho }} & =\boldsymbol{\dot{\rho }}_{r}+\boldsymbol{\omega }_{cs}\times \boldsymbol{\rho } \end{align*}
Since the angle \(\phi \) is fixed in time, hence\[ \boldsymbol{\dot{\rho }}_{r}=0 \] In this problem \[ \boldsymbol{\omega }_{cs}=\omega \mathbf{k}\] Therefore\begin{align*} \boldsymbol{\dot{\rho }} & =\boldsymbol{\omega }_{cs}\times \boldsymbol{\rho }\\ & =\omega \mathbf{k\times }\left ( -s\sin \phi \mathbf{k}+s\cos \phi \mathbf{j}\right ) \\ & =\mathbf{i}\left ( -\omega s\cos \phi \right ) \end{align*}
Therefore Eq. (1) becomes\begin{align*} d\mathbf{h}_{o} & =\left ( -s\sin \phi \mathbf{k}+s\cos \phi \mathbf{j}\right ) \times \mathbf{i}\left ( -\omega s\cos \phi \right ) \frac{m}{\pi r^{2}}dA\\ d\mathbf{h}_{o} & =\left ( \mathbf{j}\left ( \omega s^{2}\sin \phi \cos \phi \right ) +\mathbf{k}\left ( \omega s^{2}\cos \phi \cos \phi \right ) \right ) \frac{m}{\pi r^{2}}dA \end{align*}
Hence\[ \mathbf{h}_{o}=\int _{A}\left ( \mathbf{j}\left ( \omega s^{2}\sin \phi \cos \phi \right ) +\mathbf{k}\left ( \omega s^{2}\cos \phi \cos \phi \right ) \right ) \frac{m}{\pi r^{2}}dA \] Polar coordinates is used to integrate this. In polar coordinates, \(dA=sdsd\theta \) where \(s\) is the current distance from the center of the disk to the unit area, hence it goes from \(0\) to \(r\), and \(\theta \) goes from \(0\) to \(2\pi \), therefore the above becomes\begin{align*} \mathbf{h}_{o} & =\frac{m}{\pi r^{2}}\int _{\theta =0}^{\theta =2\pi }\left ( \int _{s=0}^{s=r}\left ( \mathbf{j}\left ( \omega s^{2}\sin \phi \cos \phi \right ) +\mathbf{k}\left ( \omega s^{2}\cos \phi \cos \phi \right ) \right ) sds\right ) d\theta \\ & =\frac{m}{\pi r^{2}}\int _{\theta =0}^{\theta =2\pi }\left [ \mathbf{j}\left ( \omega \frac{s^{4}}{4}\sin \phi \cos \phi \right ) +\mathbf{k}\left ( \omega \frac{s^{4}}{4}\cos \phi \cos \phi \right ) \right ] _{s=0}^{s=r}d\theta \\ & =\frac{m}{\pi r^{2}}\int _{\theta =0}^{\theta =2\pi }\mathbf{j}\left ( \omega \frac{r^{4}}{4}\sin \phi \cos \phi \right ) +\mathbf{k}\left ( \omega \frac{r^{4}}{4}\cos \phi \cos \phi \right ) d\theta \\ & =\frac{2\pi m}{r^{2}}\left ( \mathbf{j}\left ( \omega \frac{r^{4}}{4}\sin \phi \cos \phi \right ) +\mathbf{k}\left ( \omega \frac{r^{4}}{4}\cos \phi \cos \phi \right ) \right ) \\ & =\frac{\omega }{2}\pi mr^{2}\left ( \mathbf{j}\sin \phi \cos \phi +\mathbf{k}\cos \phi \cos \phi \right ) \end{align*}
Therefore\[ \mathbf{\dot{h}}_{o}=\mathbf{\dot{h}}_{o,r}+\boldsymbol{\omega }_{cs}\times \mathbf{h}_{o}\] Where\[ \mathbf{\dot{h}}_{o,r}=\frac{\alpha }{2}\pi mr^{2}\left ( \mathbf{j}\sin \phi \cos \phi +\mathbf{k}\cos \phi \cos \phi \right ) \] Hence\begin{align*} \mathbf{\dot{h}}_{o} & =\frac{\alpha }{2}\pi mr^{2}\left ( \mathbf{j}\sin \phi \cos \phi +\mathbf{k}\cos \phi \cos \phi \right ) +\omega \mathbf{k}\times \frac{\omega }{2}\pi mr^{2}\left ( \mathbf{j}\sin \phi \cos \phi +\mathbf{k}\cos \phi \cos \phi \right ) \\ & =\frac{\alpha }{2}\pi mr^{2}\left ( \mathbf{j}\sin \phi \cos \phi +\mathbf{k}\cos \phi \cos \phi \right ) -\mathbf{i}\frac{\omega ^{2}}{2}\pi mr^{2}\sin \phi \cos \phi \\ & =\mathbf{i}\frac{\omega ^{2}}{2}\pi mr^{2}\sin \phi \cos \phi +\mathbf{j}\frac{\alpha }{2}\pi mr^{2}\sin \phi \cos \phi +\mathbf{k}\frac{\alpha }{2}\pi mr^{2}\cos \phi \cos \phi \\ & =\frac{\pi mr^{2}}{2}\left ( \mathbf{i}\omega ^{2}\sin \phi \cos \phi +\mathbf{j}\alpha \sin \phi \cos \phi +\mathbf{k}\alpha \cos \phi \cos \phi \right ) \end{align*}
Therefore the torque generated by the rotating disk is\begin{align*} \mathbf{M}_{o} & =\mathbf{\dot{h}}_{o}\\ & =\frac{\pi mr^{2}}{2}\left ( \mathbf{i}\omega ^{2}\sin \phi \cos \phi +\mathbf{j}\alpha \sin \phi \cos \phi +\mathbf{k}\alpha \cos \phi \cos \phi \right ) \end{align*}
A free body diagram is now made with all the reactions on the shaft and the above found torque in order to solve for the reactions
Moment \(M_{z}\) is a torsion torque (twisting moment) and will not be considered since it does not affect shown reactions to be found. Only the moment in the \(xz\) plane (the \(M_{x}\) component) will be used to find \(A_{y},B_{y}\) and the moment in the \(yz\) plane (the \(M_{y}\) component) will be used to find \(A_{x},B_{x}\).
Taking moments at left end of the shaft, in the \(xz\) plane, gives\begin{align*} M_{x}+B_{y}L & =0\\ B_{y} & =-\frac{M_{x}}{L}=\frac{-\pi mr^{2}}{2L}\omega ^{2}\sin \phi \cos \phi \end{align*}
Taking moments at right end of the shaft, in the \(xz\) plane, gives \begin{align*} M_{x}-A_{y}L & =0\\ A_{y} & =\frac{\pi mr^{2}}{2L}\omega ^{2}\sin \phi \cos \phi \end{align*}
Taking moments at left end of the shaft, in the \(yz\) plane, gives\begin{align*} M_{y}-B_{x}L & =0\\ B_{x} & =\frac{M_{y}}{L}=\frac{\pi mr^{2}}{2L}\alpha \sin \phi \cos \phi \end{align*}
Taking moments at right end of the shaft, in the \(yz\) plane, gives\begin{align*} M_{y}+A_{x}L & =0\\ A_{x} & =\frac{-M_{y}}{L}=\frac{-\pi mr^{2}}{2L}\alpha \sin \phi \cos \phi \end{align*}
Let the body fixed coordinate system has its origin at point \(A\) and attached to the spinning disk. The following diagram shows the general configuration used to derive the equation of motion of mass \(m~\)using the relative angular momentum method.
Now all the terms needed in the above equation are found. \begin{equation} \boldsymbol{\rho }=L\sin \theta \mathbf{i}+L\cos \theta \mathbf{j} \tag{1} \end{equation} The relative angular momentum is\begin{equation} \mathbf{h}_{p}=\boldsymbol{\rho }\times m\boldsymbol{\dot{\rho }} \tag{2} \end{equation} The absolute angular acceleration of the body fixed coordinates system is\[ \boldsymbol{\omega }_{cs}=\omega \mathbf{j}\] We need to take the time derivative of \(\boldsymbol{\rho }\). Since this vector is rotating relative to the reference frame we use the standard method of adding the correction term\[ \boldsymbol{\dot{\rho }}=\boldsymbol{\dot{\rho }}_{r}+\left ( \omega \mathbf{j}\times L\sin \theta \mathbf{i}\right ) \] In the above, only the component \(L\sin \theta \mathbf{i}\) is corrected for since the body fixed axis \(\mathbf{i}\) does rotate as seen from the inertial frame of reference. The \(L\cos \theta \mathbf{j}\) does not need to be corrected for since the body fixed axis \(\mathbf{j}\) is aligned to the inertial axis \(\mathbf{J}\) all the time. Evaluating the above gives\begin{align*} \boldsymbol{\dot{\rho }} & =\left ( L\dot{\theta }\cos \theta \mathbf{i}-L\dot{\theta }\sin \theta \mathbf{j}\right ) +\left ( \omega \mathbf{j}\times L\sin \theta \mathbf{i}\right ) \\ & =L\dot{\theta }\cos \theta \mathbf{i}-L\dot{\theta }\sin \theta \mathbf{j}-L\omega \sin \theta \mathbf{k} \end{align*}
Hence \(\mathbf{h}_{p}\) from Eq. (2) becomes\begin{align*} \mathbf{h}_{p} & =\boldsymbol{\rho }\times m\boldsymbol{\dot{\rho }}\\ & =\left ( L\sin \theta \mathbf{i}+L\cos \theta \mathbf{j}\right ) \times m\left ( L\dot{\theta }\cos \theta \mathbf{i}-L\dot{\theta }\sin \theta \mathbf{j}-L\omega \sin \theta \mathbf{k}\right ) \\ & =-L^{2}\dot{\theta }\sin ^{2}\theta \mathbf{k}+\mathbf{j}L^{2}\omega \sin ^{2}\theta -\mathbf{k}L^{2}\dot{\theta }\cos ^{2}\theta -\mathbf{i}\left ( L^{2}\omega \cos \theta \sin \theta \right ) \\ & =-\mathbf{i}\left ( mL^{2}\omega \cos \theta \sin \theta \right ) +mL^{2}\omega \sin ^{2}\theta \mathbf{j}-mL^{2}\dot{\theta }\mathbf{k} \end{align*}
To make it easier to differentiate, from trig tables, let \(\cos \theta \sin \theta =\frac{1}{2}\sin \left ( 2\theta \right ) \) so that the product rule is reduced. The above becomes\[ \mathbf{h}_{p}=-\mathbf{i}\left ( \frac{1}{2}mL^{2}\omega \sin 2\theta \right ) +mL^{2}\omega \sin ^{2}\theta \mathbf{j}-mL^{2}\dot{\theta }\mathbf{k}\] The rate of change of relative angular momentum is \begin{align*} \mathbf{\dot{h}}_{p} & =\frac{d}{dt}\mathbf{h}_{p}+\left ( \omega \mathbf{j\times }\left ( -\mathbf{i}\frac{1}{2}mL^{2}\omega \sin 2\theta -mL^{2}\dot{\theta }\mathbf{k}\right ) \right ) \\ & =-\mathbf{i}\left ( \frac{1}{2}mL^{2}\omega \left ( 2\dot{\theta }\right ) \cos 2\theta \right ) +mL^{2}\omega \left ( 2\sin \theta \dot{\theta }\cos \theta \right ) \mathbf{j}-mL^{2}\ddot{\theta }\mathbf{k}\\ & +\mathbf{k}\left ( \frac{1}{2}mL^{2}\omega ^{2}\sin 2\theta \right ) -\left ( \omega mL^{2}\dot{\theta }\right ) \mathbf{i} \end{align*}
Hence\begin{align*} \mathbf{\dot{h}}_{p} & =\mathbf{i}\left ( -mL^{2}\omega \dot{\theta }\cos 2\theta -\omega mL^{2}\dot{\theta }\right ) +\left ( 2mL^{2}\dot{\theta }\omega \sin \theta \cos \theta \right ) \mathbf{j+}\left ( \frac{1}{2}mL^{2}\omega ^{2}\sin 2\theta -mL^{2}\ddot{\theta }\right ) \mathbf{k}\\ & =\mathbf{i}\left ( -mL^{2}\omega \dot{\theta }\cos 2\theta -\omega mL^{2}\dot{\theta }\right ) +\left ( mL^{2}\dot{\theta }\omega \sin \left ( 2\theta \right ) \right ) \mathbf{j+}\left ( \frac{1}{2}mL^{2}\omega ^{2}\sin 2\theta -mL^{2}\ddot{\theta }\right ) \mathbf{k} \end{align*}
Applying \(\mathbf{M}_{A}=\mathbf{\dot{h}}_{p}+m\boldsymbol{\rho }\times \mathbf{\ddot{r}}_{A}\) and since \(\mathbf{M}_{A}\) is all the applied moments around \(A\), these come from the moment applied by the torsional spring, which adds \(k_{T}\left ( \theta +\theta _{0}\right ) \) magnitude. The angle \(\theta _{0}\) is added to \(\theta \) since we are told the spring is relaxed at \(-\theta _{0}\), therefore, the total angle from the relaxed position is the absolute sum of \(\theta _{0}\) and any additional angle.
This torsional spring moment acts counter clock wise when the pendulum swings to the right as shown. Now the weight of the mass \(m~\) adds an \(mgL\sin \theta \) moment, which acts clockwise. Therefore \(\mathbf{M}_{A}=\mathbf{\dot{h}}_{p}+m\boldsymbol{\rho }\times \mathbf{\ddot{r}}_{A}\) becomes\begin{multline*} \left ( k_{T}\left ( \theta +\theta _{0}\right ) -mgL\sin \theta \right ) \mathbf{k}=\mathbf{i}\left ( -mL^{2}\omega \dot{\theta }\cos 2\theta -\omega mL^{2}\dot{\theta }\right ) +\left ( mL^{2}\dot{\theta }\omega \sin \left ( 2\theta \right ) \right ) \mathbf{j}\\ \mathbf{+}\left ( \frac{1}{2}mL^{2}\omega ^{2}\sin 2\theta -mL^{2}\ddot{\theta }\right ) \mathbf{k}+m\boldsymbol{\rho }\times \mathbf{\ddot{r}}_{A} \end{multline*} \(\mathbf{\ddot{r}}_{A}\) is the absolute acceleration of \(A\) and since \(\omega \) is constant, then only normal acceleration towards the center of disk will exist and no tangential acceleration. The normal acceleration is \(a\omega ^{2}\mathbf{i}\) in the negative \(\mathbf{i}\) direction. The above becomes\begin{align*} \left ( k_{T}\left ( \theta +\theta _{0}\right ) -mgL\sin \theta \right ) \mathbf{k} & =\mathbf{i}\left ( -mL^{2}\omega \dot{\theta }\cos 2\theta -\omega mL^{2}\dot{\theta }\right ) +\left ( mL^{2}\dot{\theta }\omega \sin \left ( 2\theta \right ) \right ) \mathbf{j}\\ & \mathbf{+}\left ( \frac{1}{2}mL^{2}\omega ^{2}\sin 2\theta -mL^{2}\ddot{\theta }\right ) \mathbf{k}+m\left ( L\sin \theta \mathbf{i}+L\cos \theta \mathbf{j}\right ) \times \left ( -a\omega ^{2}\mathbf{i}\right ) \\ & \\ & =\mathbf{i}\left ( -mL^{2}\omega \dot{\theta }\cos 2\theta -\omega mL^{2}\dot{\theta }\right ) +\left ( mL^{2}\dot{\theta }\omega \sin \left ( 2\theta \right ) \right ) \mathbf{j}\\ & \mathbf{+}\left ( \frac{1}{2}mL^{2}\omega ^{2}\sin 2\theta -mL^{2}\ddot{\theta }\right ) \mathbf{k}+mLa\omega ^{2}\cos \theta \mathbf{k} \end{align*}
Considering each component at a time, 3 scalar equations are generated one for \(\mathbf{i}\) and one for \(\mathbf{j}\) and one for \(\mathbf{k}\)\begin{align*} 0 & =-mL^{2}\omega \dot{\theta }\cos 2\theta -\omega mL^{2}\dot{\theta }\\ 0 & =mL^{2}\dot{\theta }\omega \sin \left ( 2\theta \right ) \\ k_{T}\left ( \theta +\theta _{0}\right ) -mgL\sin \theta & =\frac{1}{2}mL^{2}\omega ^{2}\sin 2\theta -mL^{2}\ddot{\theta }+mLa\omega ^{2}\cos \theta \end{align*}
The third equation (for \(\mathbf{k}\)) is the only one that contains the angular acceleration of the mass \(m\) around \(A\), hence that is the one used. Therefore the equation of motion is\begin{align*} mL^{2}\ddot{\theta }-mLa\omega ^{2}\cos \theta -mgL\sin \theta -\frac{1}{2}mL^{2}\omega ^{2}\sin 2\theta & =-k_{T}\left ( \theta +\theta _{0}\right ) \\ \ddot{\theta }-\frac{a\omega ^{2}}{L}\cos \theta -\frac{g}{L}\sin \theta -\frac{1}{2}\omega ^{2}\sin 2\theta & =-\frac{k_{T}\left ( \theta +\theta _{0}\right ) }{mL^{2}} \end{align*}
To determine \(\theta _{0}\), we are told that the spring is vertical when \(\dot{\theta }=\ddot{\theta }=0\) for constant \(\omega \). Hence from the equation of motion, letting \(\theta =0\) (since vertical position), results in\begin{align*} -\frac{a\omega ^{2}}{L} & =-\frac{k_{T}\left ( \theta _{0}\right ) }{mL^{2}}\\ \theta _{0} & =\frac{amL\omega ^{2}}{k_{T}}\qquad radian \end{align*}
Checking units to see the RHS has no units, since the LHS is radian (no units). Units of \(k_{T}\) is newton-meters per radian. Therefore\[ \frac{amL\omega ^{2}}{k_{T}}=\frac{LML\frac{1}{T^{2}}}{\frac{ML}{T^{2}}L}=1 \] Hence units are verified. The equation of motion is\[ \ddot{\theta }-\frac{a\omega ^{2}}{L}\cos \theta -\frac{g}{L}\sin \theta -\frac{1}{2}\omega ^{2}\sin 2\theta =-\frac{k_{T}\left ( \theta +\theta _{0}\right ) }{mL^{2}}\]
For small angle, \(\cos \theta =1\) and \(\sin 2\theta =2\theta \), therefore, the equation of motion becomes\begin{align*} \ddot{\theta }-\frac{a\omega ^{2}}{L}-\frac{g}{L}2\theta -\frac{1}{2}\omega ^{2}\left ( 2\theta \right ) & =-\frac{k_{T}\theta }{mL^{2}}-\frac{k_{T}\theta _{0}}{mL^{2}}\\ \ddot{\theta }-\frac{a\omega ^{2}}{L}-\frac{g}{L}2\theta -\theta \omega ^{2}+\frac{k_{T}\theta }{mL^{2}} & =-\frac{k_{T}\theta _{0}}{mL^{2}}\\ \ddot{\theta }-\frac{a\omega ^{2}}{L}+\theta \left ( \frac{k_{T}}{mL^{2}}-\omega ^{2}-\frac{2g}{L}\right ) & =-\frac{k_{T}\left ( \theta +\theta _{0}\right ) }{mL^{2}} \end{align*}
Therefore, the natural frequency is \[ \omega _{n}=\sqrt{\frac{k_{T}}{mL^{2}}-\omega ^{2}-\frac{2g}{L}}\qquad rad/\sec \] Checking units: \begin{align*} \frac{k_{T}}{mL^{2}}-\omega ^{2}-\frac{2g}{L} & =\frac{\frac{ML}{T^{2}}L}{ML^{2}}-\frac{1}{T^{2}}-\frac{L}{T^{2}L}\\ & =\frac{1}{T^{2}}-\frac{1}{T^{2}}-\frac{1}{T^{2}}\\ & =\frac{1}{T^{2}} \end{align*}
Hence \(\sqrt{\frac{1}{T^{2}}}=\frac{1}{T}\), or per second. Hence the units match to radians per second, which is the units of the natural frequency.