Let \(\boldsymbol{\omega }\) be the absolute angular velocity of the body but written using its principal unit vectors. The body in this case is the propeller which is shown above as a small bar. The \(\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\) are the body fixed principal axes of the propeller. Therefore \[ \boldsymbol{\omega }=\omega _{s}\mathbf{e}_{1}+\omega _{p}\sin \theta \mathbf{e}_{2}+\omega _{p}\cos \theta \mathbf{e}_{3}\] But \(\dot{\theta }=\omega _{s}\). This is the absolute angular velocity of the propeller itself. Hence\[ \boldsymbol{\omega }=\dot{\theta }\mathbf{e}_{1}+\omega _{p}\sin \theta \mathbf{e}_{2}+\omega _{p}\cos \theta \mathbf{e}_{3}\] We want to write everything using body principal axes to avoid taking derivatives for moments of inertial. When using \(\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\) then the moments of inertia of the propeller are constant relative to its own principal axes and also all the cross products of moments of inertia are zero, and only \(I_{1},I_{2},I_{3}\) need to be used, which simplifies the equations.\begin{align*} \boldsymbol{\dot{\omega }} & =\overset{0}{\overbrace{\ddot{\theta }\mathbf{e}_{1}}}+\omega _{p}\dot{\theta }\cos \theta \mathbf{e}_{2}-\dot{\theta }\omega _{p}\sin \theta \mathbf{e}_{3}\\ & =\omega _{p}\dot{\theta }\cos \theta \mathbf{e}_{2}-\dot{\theta }\omega _{p}\sin \theta \mathbf{e}_{3} \end{align*}
Modeling propeller as uniform slender bar\begin{align*} I_{1} & =\frac{mL^{2}}{12}\\ I_{2} & =\frac{mL^{2}}{12}\\ I_{3} & \sim 0 \end{align*}
The reference point used is the origin which is fixed on the body. Hence \[ M\mathbf{\rho }_{c}\times \mathbf{\ddot{r}}_{p}=0 \]\[ \mathbf{h}_{p}=\begin{pmatrix} I_{1} & 0 & 0\\ 0 & I_{2} & 0\\ 0 & 0 & I_{3}\end{pmatrix}\begin{pmatrix} \boldsymbol{\omega }_{1}\\ \boldsymbol{\omega }_{2}\\ \boldsymbol{\omega }_{3}\end{pmatrix} \] And \begin{align*} \mathbf{\dot{h}}_{p} & =\begin{pmatrix} \dot{h}_{1}\\ \dot{h}_{2}\\ \dot{h}_{3}\end{pmatrix} =\begin{pmatrix} I_{1}\boldsymbol{\dot{\omega }}_{1}+\boldsymbol{\omega }_{2}\boldsymbol{\omega }_{3}\left ( I_{3}-I_{2}\right ) \\ I_{2}\boldsymbol{\dot{\omega }}_{2}+\boldsymbol{\omega }_{1}\boldsymbol{\omega }_{3}\left ( I_{1}-I_{3}\right ) \\ I_{3}\boldsymbol{\dot{\omega }}_{3}+\boldsymbol{\omega }_{1}\boldsymbol{\omega }_{2}\left ( I_{2}-I_{1}\right ) \end{pmatrix} \\ & =\begin{pmatrix} \omega _{p}^{2}\sin \theta \cos \theta \left ( -\frac{mL^{2}}{12}\right ) \\ \frac{mL^{2}}{12}\omega _{p}\dot{\theta }\cos \theta +\dot{\theta }\omega _{p}\cos \theta \left ( \frac{mL^{2}}{12}\right ) \\ \dot{\theta }\omega _{p}\sin \theta \left ( \frac{mL^{2}}{12}-\frac{mL^{2}}{12}\right ) \end{pmatrix} \\ & =\begin{pmatrix} -\frac{mL^{2}}{12}\omega _{p}^{2}\sin \theta \cos \theta \\ \frac{mL^{2}}{6}\omega _{p}\dot{\theta }\cos \theta \\ 0 \end{pmatrix} \end{align*}
Hence\begin{align*} M_{o} & =\mathbf{\dot{h}}_{p}+\overset{0\text{ (fixed point)}}{\overbrace{m\mathbf{\rho }_{c}\times \mathbf{\ddot{r}}_{o}}}\\ & =\mathbf{\dot{h}}_{p} \end{align*}
When in vertical position, the angle \(\theta \) is zero, hence the dynamic moment is\[ M_{o}=\frac{mL^{2}}{6}\omega _{p}\dot{\theta }\mathbf{e}_{2}\] Converting back to \(xyz\) coordinates\[ M_{o}=\frac{mL^{2}}{4}\omega _{p}\omega _{s}\mathbf{j}\] Hence this is the torque value when \(\theta =0\) \[ \tau =\frac{mL^{2}}{6}\omega _{p}\omega _{s}\mathbf{j}\] Check units: \(\left ( ML^{2}\right ) \frac{1}{T}\frac{1}{T}=\left ( \frac{ML}{T^{2}}\right ) L=\)Force\(\times \)Length. Units agree. (I had expected the torque to be in the \(\mathbf{k}\) axes direction first. I went over this few times and do not see if I did something wrong).
Let \(\boldsymbol{\omega }\) be the absolute angular velocity of the body but written using its principal unit vectors. The body in this case is the block. The \(\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\) are the body fixed principal axes of block. Therefore \[ \boldsymbol{\omega }=\omega _{o}\cos \phi \mathbf{e}_{1}-\omega _{o}\sin \phi \mathbf{e}_{2}+\dot{\phi }\mathbf{e}_{3}\] We want to write everything using body principal axes to avoid taking derivatives for moments of inertial. When using \(\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\) then the moments of inertia of the propeller are constant relative to its own principal axes and also all the cross products of moments of inertia are zero, and only \(I_{1},I_{2},I_{3}\) need to be used, which simplifies the equations.\[ \boldsymbol{\dot{\omega }}=-\omega _{o}\dot{\phi }\sin \phi \mathbf{e}_{1}-\omega _{o}\dot{\phi }\cos \phi \mathbf{e}_{2}\] Using\begin{align*} I_{1} & =\frac{m\left ( a^{2}+b^{2}\right ) }{12}\\ I_{2} & =\frac{m\left ( b^{2}+c^{2}\right ) }{12}\\ I_{3} & =\frac{m\left ( a^{2}+c^{2}\right ) }{12} \end{align*}
The reference point used is the origin which is fixed on the body. Hence \[ M\mathbf{\rho }_{c}\times \mathbf{\ddot{r}}_{p}=0 \]\[ \mathbf{h}_{p}=\begin{pmatrix} I_{1} & 0 & 0\\ 0 & I_{2} & 0\\ 0 & 0 & I_{3}\end{pmatrix}\begin{pmatrix} \boldsymbol{\omega }_{1}\\ \boldsymbol{\omega }_{2}\\ \boldsymbol{\omega }_{3}\end{pmatrix} \] And\[ \boldsymbol{\dot{\omega }}=-\omega _{o}\dot{\phi }\sin \phi \mathbf{e}_{1}-\omega _{o}\dot{\phi }\cos \phi \mathbf{e}_{2}\]\[ \boldsymbol{\omega }=\omega _{o}\cos \phi \mathbf{e}_{1}-\omega _{o}\sin \phi \mathbf{e}_{2}+\dot{\phi }\mathbf{e}_{3}\]\begin{align*} \mathbf{\dot{h}}_{p} & =\begin{pmatrix} \dot{h}_{1}\\ \dot{h}_{2}\\ \dot{h}_{3}\end{pmatrix} =\begin{pmatrix} I_{1}\boldsymbol{\dot{\omega }}_{1}+\boldsymbol{\omega }_{2}\boldsymbol{\omega }_{3}\left ( I_{3}-I_{2}\right ) \\ I_{2}\boldsymbol{\dot{\omega }}_{2}+\boldsymbol{\omega }_{1}\boldsymbol{\omega }_{3}\left ( I_{1}-I_{3}\right ) \\ I_{3}\boldsymbol{\dot{\omega }}_{3}+\boldsymbol{\omega }_{1}\boldsymbol{\omega }_{2}\left ( I_{2}-I_{1}\right ) \end{pmatrix} \\ & =\begin{pmatrix} -I_{1}\omega _{o}\dot{\phi }\sin \phi -\dot{\phi }\omega _{o}\sin \phi \left ( I_{3}-I_{2}\right ) \\ -I_{2}\omega _{o}\dot{\phi }\cos \phi +\omega _{o}\dot{\phi }\cos \phi \left ( I_{1}-I_{3}\right ) \\ -\omega _{o}^{2}\cos \phi \sin \phi \left ( I_{2}-I_{1}\right ) \end{pmatrix} \\ & =\begin{pmatrix} \dot{\phi }\omega _{o}\sin \phi \left ( I_{2}-I_{3}-I_{1}\right ) \\ \omega _{o}\dot{\phi }\cos \phi \left ( I_{1}-I_{3}-I_{2}\right ) \\ \omega _{o}^{2}\cos \phi \sin \phi \left ( I_{2}-I_{1}\right ) \end{pmatrix} \\ & =\begin{pmatrix} p\omega _{o}\sin \phi \left ( I_{3}-I_{2}-I_{1}\right ) \\ \omega _{o}p\cos \phi \left ( I_{1}-I_{3}-I_{2}\right ) \\ \omega _{o}^{2}\cos \phi \sin \phi \left ( I_{2}-I_{1}\right ) \end{pmatrix} \end{align*}
Hence\begin{align*} \mathbf{M}_{o} & =\mathbf{\dot{h}}_{p}+\overset{0\text{ (fixed point)}}{\overbrace{m\mathbf{\rho }_{c}\times \mathbf{\ddot{r}}_{o}}}\\ & =\mathbf{\dot{h}}_{p} \end{align*}
Convert back to \(xyz\) coordinates using \begin{align*} \mathbf{e}_{3} & =\mathbf{k}\\ \mathbf{e}_{2} & =\mathbf{j}\cos \phi -\mathbf{i}\sin \phi \\ \mathbf{e}_{1} & =\mathbf{i}\cos \phi +\mathbf{j}\sin \phi \end{align*}
Hence\begin{align*} \mathbf{M}_{o} & =\left [ p\omega _{o}\sin \phi \left ( I_{2}-I_{3}-I_{1}\right ) \right ] \left ( \mathbf{i}\cos \phi +\mathbf{j}\sin \phi \right ) \\ & +\left [ \omega _{o}p\cos \phi \left ( I_{1}-I_{3}-I_{2}\right ) \right ] \left ( \mathbf{j}\cos \phi -\mathbf{i}\sin \phi \right ) \\ & +\omega _{o}^{2}\cos \phi \sin \phi \left ( I_{2}-I_{1}\right ) \mathbf{k} \end{align*}
Or\begin{align*} \mathbf{M}_{o} & =\mathbf{i}\left [ p\omega _{o}\sin \phi \cos \phi \left ( I_{2}-I_{3}-I_{1}\right ) -\omega _{o}p\cos \phi \sin \phi \left ( I_{1}-I_{3}-I_{2}\right ) \right ] \\ & +\mathbf{j}\left [ -p\omega _{o}\sin \phi \left ( I_{3}-I_{2}-I_{1}\right ) +\omega _{o}p\cos \phi \left ( I_{1}-I_{3}-I_{2}\right ) \right ] \\ & +\omega _{o}^{2}\cos \phi \sin \phi \left ( I_{2}-I_{1}\right ) \mathbf{k} \end{align*}
Or\begin{align*} \mathbf{M}_{o} & =2\left ( I_{2}-I_{1}\right ) p\omega _{o}\sin \phi \cos \phi \mathbf{i}\\ & +p\omega _{o}\left ( -\sin \phi \left ( I_{3}-I_{2}-I_{1}\right ) +\cos \phi \left ( I_{1}-I_{3}-I_{2}\right ) \right ) \mathbf{j}\\ & +\omega _{o}^{2}\cos \phi \sin \phi \left ( I_{2}-I_{1}\right ) \mathbf{k} \end{align*}
So the torque \(M_{t}\) is the \(\mathbf{i}\) component above, Hence \begin{align*} M_{t} & =2\left ( I_{2}-I_{1}\right ) p\omega _{o}\sin \phi \cos \phi \mathbf{i}\\ & =2\left ( \frac{m\left ( b^{2}+c^{2}\right ) }{12}-\frac{m\left ( a^{2}+b^{2}\right ) }{12}\right ) p\omega _{o}\sin \phi \cos \phi \mathbf{i}\\ & =\frac{1}{6}m\left ( c^{2}-a^{2}\right ) p\omega _{o}\sin \phi \cos \phi \mathbf{i} \end{align*}