Let \(A\) be the reference point (the point the moments will be taken about ). By using a body axes which is also a principal body axes at point \(A\) we can use Euler equations for the body fixed coordinates.
The absolute angular velocity of the reference frame is \(\mathbf{\omega }_{cs}=\omega \mathbf{k}\) and the body absolute angular velocity is \(\mathbf{\Omega }=\omega \mathbf{k}-\dot{\theta }\mathbf{i}\). This is now written in body fixed coordinates \(e_{1},e_{2},e_{3}\), hence\[ \mathbf{\Omega }=\omega \left ( \cos \theta \mathbf{e}_{3}-\sin \theta \mathbf{e}_{2}\right ) -\dot{\theta }\mathbf{e}_{1}\]
\begin{align*} \boldsymbol{\Omega }_{1} & =-\dot{\theta }\mathbf{e}_{1}\\ \boldsymbol{\Omega }_{2} & =-\sin \theta \mathbf{e}_{2}\\ \boldsymbol{\Omega }_{3} & =\cos \theta \mathbf{e}_{3} \end{align*}
And\begin{align*} \boldsymbol{\dot{\Omega }}_{1} & =-\ddot{\theta }\mathbf{e}_{1}\\ \boldsymbol{\dot{\Omega }}_{2} & =-\dot{\theta }\cos \theta \mathbf{e}_{1}\\ \boldsymbol{\dot{\Omega }}_{3} & =-\dot{\theta }\sin \theta \mathbf{e}_{3} \end{align*}
And\begin{align*} I_{1} & =\frac{ml^{2}}{3}\\ I_{2} & =\frac{ml^{2}}{3}\\ I_{3} & \sim 0 \end{align*}
Hence \[ \mathbf{h}_{A}=\begin{pmatrix} I_{1} & 0 & 0\\ 0 & I_{2} & 0\\ 0 & 0 & I_{3}\end{pmatrix}\begin{pmatrix} \boldsymbol{\Omega }_{1}\\ \boldsymbol{\Omega }_{2}\\ \boldsymbol{\Omega }_{3}\end{pmatrix} \] The rate of change of the relative angular momentum of the beam using Euler equations is\[ \mathbf{\dot{h}}_{A}=\begin{pmatrix} \dot{h}_{1}\\ \dot{h}_{2}\\ \dot{h}_{3}\end{pmatrix} =\begin{pmatrix} I_{1}\dot{\Omega }_{1}+\Omega _{2}\Omega _{3}\left ( I_{3}-I_{2}\right ) \\ I_{2}\dot{\Omega }_{2}+\Omega _{1}\Omega _{3}\left ( I_{1}-I_{3}\right ) \\ I_{3}\dot{\Omega }_{3}+\Omega _{1}\Omega _{2}\left ( I_{2}-I_{1}\right ) \end{pmatrix} \] Therefore, the moment needed to move the beam with the angular velocity specified is given by\[ \mathbf{M}_{A}=\mathbf{\dot{h}}_{A}+m\mathbf{\rho }_{c}\times \mathbf{\ddot{r}}_{A}\] Where \(\mathbf{\rho }_{c}\) is a vector from \(A\) to mass center of bar given by \(\frac{l}{2}\mathbf{e}_{3}\)and \(\mathbf{\ddot{r}}_{A}\) is the absolute angular acceleration of point \(A.\) Since the \(xyz\) rotates with constant angular velocity \(\omega \), then point \(A\) will not be accelerating in the tangential direction, but will have an acceleration inwards towards \(O\) which is \(\mathbf{\ddot{r}}_{A}=-a\omega ^{2}\mathbf{j}=-a\omega ^{2}\left ( \sin \theta \mathbf{e}_{3}+\cos \theta \mathbf{e}_{2}\right ) \), hence\begin{align*} m\mathbf{\rho }_{c}\times \mathbf{\ddot{r}}_{p} & =-m\frac{l}{2}\mathbf{e}_{3}\times a\omega ^{2}\left ( \sin \theta \mathbf{e}_{3}+\cos \theta \mathbf{e}_{2}\right ) \\ & =-ma\omega ^{2}\frac{l}{2}\left ( \mathbf{e}_{3}\times \left ( \sin \theta \mathbf{e}_{3}+\cos \theta \mathbf{e}_{2}\right ) \right ) \\ & =-ma\omega ^{2}\frac{l}{2}\left ( -\cos \theta \mathbf{e}_{1}\right ) \\ & =\frac{l}{2}ma\omega ^{2}\cos \theta \mathbf{e}_{1} \end{align*}
Therefore, \begin{align*} M_{1} & =I_{1}\dot{\Omega }_{1}+\Omega _{2}\Omega _{3}\left ( I_{3}-I_{2}\right ) +\frac{l}{2}ma\omega ^{2}\cos \theta \\ M_{2} & =I_{2}\dot{\Omega }_{2}+\Omega _{1}\Omega _{3}\left ( I_{1}-I_{3}\right ) \\ M_{3} & =I_{3}\dot{\Omega }_{3}+\Omega _{1}\Omega _{2}\left ( I_{2}-I_{1}\right ) \end{align*}
Convert back to \(xyz\) using\begin{align*} M_{x} & =M_{1}\\ M_{y} & =M_{2}\cos \theta +M_{3}\sin \theta \\ M_{z} & =M_{3}\cos \theta -M_{2}\sin \theta \end{align*}
The above gives the dynamic moment, due to rotation of bar, about \(A\) expressed in \(xyz\) coordinates. They will be written in full and simplified in order to obtain the solution. Using\begin{align*} \boldsymbol{\Omega }_{1} & =-\dot{\theta }\mathbf{e}_{1}\\ \boldsymbol{\Omega }_{2} & =-\sin \theta \mathbf{e}_{2}\\ \boldsymbol{\Omega }_{3} & =\cos \theta \mathbf{e}_{3} \end{align*}
And\begin{align*} \boldsymbol{\dot{\Omega }}_{1} & =-\ddot{\theta }\mathbf{e}_{1}\\ \boldsymbol{\dot{\Omega }}_{2} & =-\dot{\theta }\cos \theta \mathbf{e}_{1}\\ \boldsymbol{\dot{\Omega }}_{3} & =-\dot{\theta }\sin \theta \mathbf{e}_{3} \end{align*}
Then, converting back to \(xyz\) coordinates\begin{align*} M_{x} & =I_{1}\dot{\Omega }_{1}+\Omega _{2}\Omega _{3}\left ( I_{3}-I_{2}\right ) +\frac{l}{2}ma\omega ^{2}\cos \theta \\ & =-\frac{ml^{2}}{3}\ddot{\theta }-\sin \theta \cos \theta \left ( 0-\frac{ml^{2}}{3}\right ) +\frac{l}{2}ma\omega ^{2}\cos \theta \\ & =-\frac{ml^{2}}{3}\ddot{\theta }+\frac{ml^{2}}{3}\sin \theta \cos \theta +\frac{l}{2}ma\omega ^{2}\cos \theta \end{align*}
And
\begin{align*} M_{y} & =M_{2}\cos \theta +M_{3}\sin \theta \\ & =\left ( I_{2}\dot{\Omega }_{2}+\Omega _{1}\Omega _{3}\left ( I_{1}-I_{3}\right ) \right ) \cos \theta +\overbrace{\left ( I_{3}\dot{\Omega }_{3}+\Omega _{1}\Omega _{2}\left ( I_{2}-I_{1}\right ) \right ) \sin \theta }^{vanish}\\ & =\left [ -\frac{ml^{2}}{3}\dot{\theta }\cos \theta -\dot{\theta }\cos \theta \left ( \frac{ml^{2}}{3}\right ) \right ] \cos \theta \\ & =-\frac{ml^{2}}{3}\dot{\theta }\cos ^{2}\theta -\dot{\theta }\cos ^{2}\theta \left ( \frac{ml^{2}}{3}\right ) \\ & =-\frac{2}{3}ml^{2}\dot{\theta }\cos ^{2}\theta \end{align*}
And\begin{align*} M_{z} & =M_{3}\cos \theta -M_{2}\sin \theta \\ & =\overbrace{\left [ I_{3}\dot{\Omega }_{3}+\Omega _{1}\Omega _{2}\left ( I_{2}-I_{1}\right ) \right ] \cos \theta }^{vanish}-\left [ -\frac{ml^{2}}{3}\dot{\theta }\cos \theta -\dot{\theta }\cos \theta \frac{ml^{2}}{3}\right ] \sin \theta \\ & =\frac{ml^{2}}{3}\dot{\theta }\cos \theta \sin \theta +\dot{\theta }\cos \theta \sin \theta \frac{ml^{2}}{3}\\ & =\frac{2}{3}ml^{2}\dot{\theta }\cos \theta \sin \theta \end{align*}
Since the problem asks to find the rotational equation of motion around \(A\) as shown, then only \(M_{x}\) will be used. A free body diagram is used to find the external torque around \(A\)
Hence \begin{align*} -mg\frac{l}{2}\sin \theta +kl^{2}\sin \theta & =M_{x}\\ -mg\frac{l}{2}\sin \theta +kl^{2}\sin \theta & =-\frac{ml^{2}}{3}\ddot{\theta }+\frac{ml^{2}}{3}\sin \theta \cos \theta +\frac{l}{2}ma\omega ^{2}\cos \theta \end{align*}
For small angle \(\sin \theta \rightarrow \theta \) and \(\cos \theta \rightarrow 1\), hence\begin{align*} -mg\frac{l}{2}\theta +kl^{2}\theta & =-\frac{ml^{2}}{3}\ddot{\theta }+\frac{ml^{2}}{3}\theta +\frac{l}{2}ma\omega ^{2}\\ \frac{ml^{2}}{3}\ddot{\theta }-\frac{ml^{2}}{3}\theta -mg\frac{l}{2}\theta +kl^{2}\theta & =\frac{l}{2}ma\omega ^{2}\\ \frac{ml^{2}}{3}\ddot{\theta }+\left ( kl^{2}-\frac{ml^{2}}{3}-mg\frac{l}{2}\right ) \theta & =\frac{l}{2}ma\omega ^{2}\\ \ddot{\theta }+\left ( \frac{3k}{m}-1-\frac{3}{2}\frac{g}{l}\right ) \theta & =\frac{l}{2}ma\omega ^{2} \end{align*}
This is the equation of motion for rotation for small angles.
We need to find \(\mathbf{F}_{weld}\), which represent reaction at the hinge \(A\). Balance of external forces at \(A\) gives\[ \mathbf{F}_{weld}-mg\mathbf{k}-kl\sin \theta \mathbf{j}=m\mathbf{a}_{cg}\] Where \(\mathbf{a}_{cg}\) is the acceleration of center of mass of bar. Using
\begin{align*} \boldsymbol{\rho } & =\left ( a+\frac{l}{2}\sin \theta \right ) \mathbf{j+}\frac{l}{2}\cos \theta \mathbf{k}\\ \boldsymbol{\dot{\rho }}_{r} & =\left ( \frac{l}{2}\dot{\theta }\cos \theta \right ) \mathbf{j-}\frac{l}{2}\dot{\theta }\sin \theta \mathbf{k}\\ \boldsymbol{\ddot{\rho }}_{r} & =\frac{l}{2}\left ( \ddot{\theta }\cos \theta -\dot{\theta }^{2}\sin \theta \right ) \mathbf{j-}\frac{l}{2}\left ( \ddot{\theta }\sin \theta +\dot{\theta }^{2}\cos \theta \right ) \mathbf{k}\\ \boldsymbol{\omega } & =\omega \mathbf{k}\\ \boldsymbol{\dot{\omega }} & =\dot{\omega }\mathbf{k}\\ \boldsymbol{\omega }\times \boldsymbol{\dot{\rho }}_{r} & =\omega \mathbf{k}\times \left [ \left ( \frac{l}{2}\dot{\theta }\cos \theta \right ) \mathbf{j-}\frac{l}{2}\dot{\theta }\sin \theta \mathbf{k}\right ] \\ & =-\omega \frac{l}{2}\dot{\theta }\cos \theta \mathbf{i}\\ \boldsymbol{\dot{\omega }}\times \boldsymbol{\rho } & =\dot{\omega }\mathbf{k}\times \left [ \left ( a+\frac{l}{2}\sin \theta \right ) \mathbf{j+}\frac{l}{2}\cos \theta \mathbf{k}\right ] \\ & =-\dot{\omega }\left ( a+\frac{l}{2}\sin \theta \right ) \mathbf{i}\\ \boldsymbol{\omega }\times \boldsymbol{\rho } & \boldsymbol{=}-\omega \left ( a+\frac{l}{2}\sin \theta \right ) \mathbf{i}\\ \boldsymbol{\omega }\times \left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) & =\omega \mathbf{k}\times -\left ( \omega \left ( a+\frac{l}{2}\sin \theta \right ) \mathbf{i}\right ) \\ & =-\omega ^{2}\left ( a+\frac{l}{2}\sin \theta \right ) \mathbf{j}\\ \mathbf{\ddot{R}} & =0 \end{align*}
Hence\begin{align*} \mathbf{a}_{cg} & =\mathbf{\ddot{R}}+\boldsymbol{\ddot{\rho }}_{r}+2\left ( \boldsymbol{\omega }\times \boldsymbol{\dot{\rho }}_{r}\right ) +\boldsymbol{\dot{\omega }}\times \boldsymbol{\rho }+\boldsymbol{\omega }\times \left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) \\ & =\frac{l}{2}\left ( \ddot{\theta }\cos \theta -\dot{\theta }^{2}\sin \theta \right ) \mathbf{j-}\frac{l}{2}\left ( \ddot{\theta }\sin \theta +\dot{\theta }^{2}\cos \theta \right ) \mathbf{k}\\ & +2\left ( -\omega \frac{l}{2}\dot{\theta }\cos \theta \mathbf{i}\right ) -\dot{\omega }\left ( a+\frac{l}{2}\sin \theta \right ) \mathbf{i}-\omega ^{2}\left ( a+\frac{l}{2}\sin \theta \right ) \mathbf{j} \end{align*}
Hence\begin{align*} \mathbf{a}_{cg} & =\mathbf{\ddot{R}}+\boldsymbol{\ddot{\rho }}_{r}+2\left ( \boldsymbol{\omega }\times \boldsymbol{\dot{\rho }}_{r}\right ) +\boldsymbol{\dot{\omega }}\times \boldsymbol{\rho }+\boldsymbol{\omega }\times \left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) \\ & =\mathbf{i}\left ( -\omega l\dot{\theta }\cos \theta -\dot{\omega }\left ( a+\frac{l}{2}\sin \theta \right ) \right ) \\ & +\mathbf{j}\left ( \frac{l}{2}\left ( \ddot{\theta }\cos \theta -\dot{\theta }^{2}\sin \theta \right ) -\omega ^{2}\left ( a+\frac{l}{2}\sin \theta \right ) \right ) \\ & \mathbf{-}\frac{l}{2}\left ( \ddot{\theta }\sin \theta +\dot{\theta }^{2}\cos \theta \right ) \mathbf{k} \end{align*}
Hence from\[ \mathbf{F}_{weld}-mg\mathbf{k}-kl\sin \theta \mathbf{j}=m\mathbf{a}_{cg}\] We can find \(\mathbf{F}_{weld}\)\begin{align*} F_{z} & =mg-m\frac{l}{2}\left ( \ddot{\theta }\sin \theta +\dot{\theta }^{2}\cos \theta \right ) \\ F_{y} & =kl\sin \theta +m\frac{l}{2}\left ( \ddot{\theta }\cos \theta -\dot{\theta }^{2}\sin \theta \right ) -m\omega ^{2}\left ( a+\frac{l}{2}\sin \theta \right ) \\ F_{x} & =-m\omega l\dot{\theta }\cos \theta -m\dot{\omega }\left ( a+\frac{l}{2}\sin \theta \right ) \end{align*}
For small angle\begin{align*} F_{z} & =mg-m\frac{l}{2}\left ( \ddot{\theta }\theta +\dot{\theta }^{2}\right ) \\ F_{y} & =kl\theta +m\frac{l}{2}\left ( \ddot{\theta }-\dot{\theta }^{2}\theta \right ) -m\omega ^{2}\left ( a+\frac{l}{2}\theta \right ) \\ F_{x} & =-m\omega l\dot{\theta }-m\dot{\omega }\left ( a+\frac{l}{2}\theta \right ) \end{align*}
Sometimes \(\dot{\theta }^{2}\) can be approximated to zero for small angle. If this is allowed, then the above simplifies to\begin{align*} F_{z} & =mg-m\frac{l}{2}\ddot{\theta }\theta \\ F_{y} & =kl\theta +m\frac{l}{2}\ddot{\theta }-m\omega ^{2}\left ( a+\frac{l}{2}\theta \right ) \\ F_{x} & =-m\omega l\dot{\theta }-m\dot{\omega }\left ( a+\frac{l}{2}\theta \right ) \end{align*}
Since \(\ddot{\theta }\) has been found above, all reactions at joint \(A\) can now be found.
Let \(C\) be the reference point (the point the moments will be taken about ). It is also the center of mass of the rod.
The absolute angular velocity of the reference frame is \(\mathbf{\omega }_{cs}=N\mathbf{k}\) and the body absolute angular velocity is \(\mathbf{\Omega }=N\mathbf{k}+\dot{\theta }\mathbf{i}\). This is now written in body fixed coordinates \(e_{1},e_{2},e_{3}\), hence\[ \mathbf{\Omega }=N\left ( \sin \theta \mathbf{e}_{1}+\cos \theta \mathbf{e}_{2}\right ) +\dot{\theta }\mathbf{e}_{3}\] Therefore\begin{align*} \Omega _{1} & =N\sin \theta \\ \Omega _{2} & =N\cos \theta \\ \Omega _{3} & =\dot{\theta } \end{align*}
And\begin{align*} \dot{\Omega }_{1} & =N\dot{\theta }\cos \theta \\ \dot{\Omega }_{2} & =-N\dot{\theta }\sin \theta \\ \dot{\Omega }_{3} & =\ddot{\theta }=0 \end{align*}
And\begin{align*} I_{1} & \sim 0\\ I_{2} & =\frac{ml^{2}}{12}\\ I_{3} & =\frac{ml^{2}}{12} \end{align*}
Hence \[ \mathbf{h}_{c}=\begin{pmatrix} I_{1} & 0 & 0\\ 0 & I_{2} & 0\\ 0 & 0 & I_{3}\end{pmatrix}\begin{pmatrix} \boldsymbol{\Omega }_{1}\\ \boldsymbol{\Omega }_{2}\\ \boldsymbol{\Omega }_{3}\end{pmatrix} \] The rate of change of the relative angular momentum of the beam using Euler equations is\[ \mathbf{\dot{h}}_{c}=\begin{pmatrix} \dot{h}_{1}\\ \dot{h}_{2}\\ \dot{h}_{3}\end{pmatrix} =\begin{pmatrix} I_{1}\dot{\Omega }_{1}+\Omega _{2}\Omega _{3}\left ( I_{3}-I_{2}\right ) \\ I_{2}\dot{\Omega }_{2}+\Omega _{1}\Omega _{3}\left ( I_{1}-I_{3}\right ) \\ I_{3}\dot{\Omega }_{3}+\Omega _{1}\Omega _{2}\left ( I_{2}-I_{1}\right ) \end{pmatrix} \] Therefore, the moment needed to move the beam with the angular velocity specified is given by\[ \mathbf{M}_{c}=\mathbf{\dot{h}}_{A}+m\mathbf{\rho }_{c}\times \mathbf{\ddot{r}}_{c}\] Since the reference point is at the mass center of the rotating body, then \(\mathbf{\rho }_{c}=0\,\ \)Therefore, \begin{align*} M_{1} & =I_{1}\dot{\Omega }_{1}+\Omega _{2}\Omega _{3}\left ( I_{3}-I_{2}\right ) \\ M_{2} & =I_{2}\dot{\Omega }_{2}+\Omega _{1}\Omega _{3}\left ( I_{1}-I_{3}\right ) \\ M_{3} & =I_{3}\dot{\Omega }_{3}+\Omega _{1}\Omega _{2}\left ( I_{2}-I_{1}\right ) \end{align*}
Convert back to \(x^{\prime }y^{\prime }z^{\prime }\) using\begin{align*} M_{x^{\prime }} & =M_{3}\\ M_{y^{\prime }} & =M_{1}\cos \theta -M_{2}\sin \theta \\ M_{z^{\prime }} & =M_{1}\sin \theta +M_{2}\cos \theta \end{align*}
Hence\begin{align*} M_{x^{\prime }} & =I_{3}\dot{\Omega }_{3}+\Omega _{1}\Omega _{2}\left ( I_{2}-I_{1}\right ) \\ & =N^{2}\sin \theta \cos \theta \frac{ml^{2}}{12}\\ & \\ M_{y^{\prime }} & =M_{1}\cos \theta -M_{2}\sin \theta \\ & =\overbrace{\left [ I_{1}\dot{\Omega }_{1}+\Omega _{2}\Omega _{3}\left ( I_{3}-I_{2}\right ) \right ] }^{vanish}\cos \theta -\left [ I_{2}\dot{\Omega }_{2}+\Omega _{1}\Omega _{3}\left ( I_{1}-I_{3}\right ) \right ] \sin \theta \\ & =-\left [ -\frac{ml^{2}}{12}N\dot{\theta }\sin \theta +\dot{\theta }N\sin \theta \left ( 0-\frac{ml^{2}}{12}\right ) \right ] \sin \theta \\ & =\frac{ml^{2}}{12}N\dot{\theta }\sin ^{2}\theta +\dot{\theta }N\sin ^{2}\theta \frac{ml^{2}}{12}\\ & =\frac{ml^{2}}{6}N\dot{\theta }\sin ^{2}\theta \\ & \\ M_{z^{\prime }} & =M_{1}\sin \theta +M_{2}\cos \theta \\ & =\left [ I_{1}\dot{\Omega }_{1}+\Omega _{2}\Omega _{3}\left ( I_{3}-I_{2}\right ) \right ] \sin \theta +\left [ I_{2}\dot{\Omega }_{2}+\Omega _{1}\Omega _{3}\left ( I_{1}-I_{3}\right ) \right ] \cos \theta \\ & =\dot{\theta }N\cos \theta \sin \theta \left ( \frac{ml^{2}}{12}\right ) +\left [ -\frac{ml^{2}}{12}N\dot{\theta }\sin \theta +\dot{\theta }N\sin \theta \left ( 0-\frac{ml^{2}}{12}\right ) \right ] \cos \theta \\ & =\dot{\theta }N\cos \theta \sin \theta \left ( \frac{ml^{2}}{12}\right ) -\frac{ml^{2}}{12}N\dot{\theta }\sin \theta \cos \theta -\dot{\theta }N\sin \theta \cos \theta \frac{ml^{2}}{12}\\ & =-\frac{ml^{2}}{12}N\dot{\theta }\sin \theta \cos \theta \end{align*}
The above is the components of the resultant moment at \(C\) to sustain this motion.
The bar’s center of mass does not move in space. Hence there is no linear acceleration associated with the bar translation. Therefore, we can set up the free body diagram now and solve for the reactions as follows
Dynamic loads balance with external forces
To find \(\mathbf{F}_{B}\), Taking moments at \(D\) \begin{align*} -2q\mathbf{i}\times \mathbf{F}_{B}+\left ( -q\mathbf{i}\right ) \times \left ( -mg\mathbf{k}\right ) & =\mathbf{M}_{c}\\ -2q\mathbf{i}\times \left ( F_{x}\mathbf{i+}F_{y}\mathbf{j+}F_{z}\mathbf{k}\right ) -mgq\mathbf{j} & =\mathbf{M}_{c}\\ \mathbf{k}\left ( -2qF_{y}\right ) -\mathbf{j}\left ( -2qF_{z}\right ) -mgq\mathbf{j} & =\mathbf{M}_{c}\\ -2qF_{y}\mathbf{k}+2qF_{z}\mathbf{j}-mgq\mathbf{j} & =\mathbf{M}_{c} \end{align*}
For vertical reactions only, hence need to find \(F_{z}\)\begin{align*} 2qF_{z}-mgq & =\frac{ml^{2}}{6}N\dot{\theta }\sin ^{2}\theta \\ 2qF_{z} & =\frac{ml^{2}}{6}N\dot{\theta }\sin ^{2}\theta +mgq\\ F_{z} & =\frac{ml^{2}}{12q}N\dot{\theta }\sin ^{2}\theta +\frac{mg}{2} \end{align*}
The force in the bearing \(F_{z}\) is positive at \(B\). hence upwards.
To find \(F_{z}\) at \(D.\) taking moments at \(B\)\begin{align*} 2q\mathbf{i}\times \mathbf{F}_{D}+\left ( q\mathbf{i}\right ) \times \left ( -mg\mathbf{k}\right ) & =\mathbf{M}_{c}\\ 2q\mathbf{i}\times \left ( F_{x}\mathbf{i+}F_{y}\mathbf{j+}F_{z}\mathbf{k}\right ) +mgq\mathbf{j} & =\mathbf{M}_{c}\\ \mathbf{k}\left ( 2qF_{y}\right ) -\mathbf{j}\left ( 2qF_{z}\right ) +mgq\mathbf{j} & =\mathbf{M}_{c}\\ 2qF_{y}\mathbf{k}-2qF_{z}\mathbf{j}+mgq\mathbf{j} & =\mathbf{M}_{c} \end{align*}
For vertical reactions only, hence need to find \(F_{z}\)\begin{align*} -2qF_{z}+mgq & =\frac{ml^{2}}{6}N\dot{\theta }\sin ^{2}\theta \\ -2qF_{z} & =\frac{ml^{2}}{6}N\dot{\theta }\sin ^{2}\theta -mgq\\ F_{z} & =-\frac{ml^{2}}{12q}N\dot{\theta }\sin ^{2}\theta +\frac{mg}{2} \end{align*}
The force in the bearing \(F_{z}\) when \(t=0\) is positive. but it can become negative. It depends if \(\frac{ml^{2}}{12q}N\dot{\theta }\sin ^{2}\theta \) is bigger or smaller than \(\frac{mg}{2}\)