5.3 Problem 7.2 part (e) in Textbook

Starting with

\begin{align} q^{\prime }\left ( t\right ) +\frac {\left ( I_{1}-I_{3}\right ) }{I_{2}}\Omega \ r\left ( t\right ) & =0\tag {1}\\ r^{\prime }\left ( t\right ) +\frac {\left ( I_{2}-I_{1}\right ) }{I_{3}}\Omega \ q\left ( t\right ) & =0\tag {2}\end{align}

To decouple the ODE’s, take derivatives and substituting back, we find

\begin{align} q^{\prime \prime }\left ( t\right ) +k\ q\left ( t\right ) & =0\tag {3}\\ r^{\prime \prime }\left ( t\right ) +k\ r\left ( t\right ) & =0\tag {4}\end{align}

Where \(k=\frac {\left ( I_{1}-I_{3}\right ) \left ( I_{2}-I_{2}\right ) }{I_{2}I_{3}}\Omega ^{2}\). The solution to the above is (for stability \(k>0\) )

\begin{align} q\left ( t\right ) & =A\cos \sqrt {k}t+B\sin \sqrt {k}t\tag {5}\\ r\left ( t\right ) & =C\cos \sqrt {k}t+D\sin \sqrt {k}t\tag {6}\end{align}

Now, \(q\left ( 0\right ) =\frac {\text {Imp}}{I_{yy}}\), hence \(A=\frac {\text {Imp}}{I_{yy}}\) . To find \(B\) take derivative

\begin{equation} q^{\prime }\left ( t\right ) =-\sqrt {k}\frac {\text {Imp}}{I_{yy}}\sin \sqrt {k}t+B\sqrt {k}\cos \sqrt {k}t\tag {7}\end{equation}

But at \(t=0\) then we go back and use Eq. (1) to find \(q^{\prime }\left ( 0\right ) \), and equate the result to the above at \(t=0\). Eq (1), at \(t=0\), gives

\[ q^{\prime }\left ( 0\right ) +\frac {\left ( I_{1}-I_{3}\right ) }{I_{2}}\Omega \ r\left ( 0\right ) =0 \]

But \(r\left ( 0\right ) =0\,\), hence \(q^{\prime }\left ( 0\right ) =0\), and so this results in \(B=0\) in Eq.(7), hence the solution for \(q\left ( t\right ) \) is

\[ q\left ( t\right ) =\frac {\text {Imp}}{I_{yy}}\cos \sqrt {k}t \]

We do the same for \(r\left ( t\right ) .\) From Eq. (6) we find that \(C=0\) since \(r\left ( 0\right ) =0\), so now \(r\left ( t\right ) \) reduces to

\[ r\left ( t\right ) =D\sin \sqrt {k}t \]

Hence

\begin{equation} r^{\prime }\left ( t\right ) =\sqrt {k}D\cos \sqrt {k}t\tag {8}\end{equation}

To find \(D\), then we go back and use Eq. (2) to find \(r^{\prime }\left ( 0\right ) \), and equate the result to the above at \(t=0\). Eq (2), at \(t=0\), gives

\[ r^{\prime }\left ( 0\right ) +\frac {\left ( I_{2}-I_{1}\right ) }{I_{3}}\Omega \ q\left ( 0\right ) =0 \]

But \(q\left ( 0\right ) =\frac {\text {Imp}}{I_{yy}}\), hence

\[ r^{\prime }\left ( 0\right ) =-\frac {\left ( I_{2}-I_{1}\right ) }{I_{3}}\Omega \ \frac {\text {Imp}}{I_{yy}}\]

Therefore, equate the above to Eq. (8) evaluated at \(t=0\,\ \)gives

\[ D=-\frac {\left ( I_{2}-I_{1}\right ) }{I_{3}\sqrt {k}}\Omega \ \frac {\text {Imp}}{I_{yy}}\]

Hence Eq. (6) becomes

\[ r\left ( t\right ) =\left ( -\frac {\left ( I_{2}-I_{1}\right ) }{I_{3}\sqrt {k}}\Omega \ \frac {\text {Imp}}{I_{yy}}\right ) \sin \sqrt {k}t \]

so \(r\left ( t\right ) \) is sinusoidal as well.