28.2: Free Rotation of a Symmetric Top Using Euler’s Equations
- Page ID
- 30676
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)This is a problem we’ve already solved, using Lagrangian methods and Euler angles, but it’s worth seeing just how easy it is using Euler’s equations.
For \(I_{1}=I_{2}\), the third equation gives immediately \(\Omega_{3}=\text { constant. }\).
Then, writing for convenience
\begin{equation}
\Omega_{3}\left(I_{3}-I_{1}\right) / I_{1}=\omega
\end{equation}
the first two equations are
\begin{equation}
\dot{\Omega}_{1}=-\omega \Omega_{2}, \quad \dot{\Omega}_{2}=\omega \Omega_{1}
\end{equation}
These equations can be combined to give
\begin{equation}
d\left(\Omega_{1}+i \Omega_{2}\right) / d t=i \omega\left(\Omega_{1}+i \Omega_{2}\right), \text { sо }\left(\Omega_{1}+i \Omega_{2}\right)=A e^{i \omega t}
\end{equation}
That is, \(\left(\Omega_{1}, \Omega_{2}\right)\) moves around a circle centered at the origin with constant angular velocity. So \(\begin{equation}\Omega_{1}^{2}+\Omega_{2}^{2}=|A|^{2}\end{equation}\) stays constant, and \(\Omega_{3}\) is constant, the angular velocity vector has constant length and rotates steadily about the axis \(x_{3}\).
From
\begin{equation}
L_{1}=I_{1} \Omega_{1}, L_{2}=I_{2} \Omega_{2}, L_{3}=I_{3} \Omega_{3}
\end{equation}
it follows that the angular momentum vector also precesses at a steady rate about \(x_{3}\). This is, remember, in the body frame—we know that in the fixed space frame, the angular momentum vector is constant! It follows that, as viewed from the outside, the \(x_{3}\) axis precesses around the fixed angular momentum vector at a steady rate.
Of course, the rate is the same as that found using Euler’s angles, recall from the previous lecture that
\begin{equation}
\vec{L}=\left(I_{1} \Omega_{1}, I_{1} \Omega_{2}, I_{3} \Omega_{3}\right)=\left(I_{1} \dot{\theta}, \quad I_{1} \dot{\phi} \sin \theta, \quad I_{3}(\dot{\phi} \cos \theta+\dot{\psi})\right)
\end{equation}
so in precession
\begin{equation}
L_{3}=L \cos \theta=I_{3}(\dot{\phi} \cos \theta+\dot{\psi}) \text { and } \dot{\phi}=L / I_{1}
\end{equation}
so
\begin{equation}
\dot{\psi}=L \cos \theta\left(\dfrac{1}{I_{3}}-\dfrac{1}{I_{1}}\right)=-\Omega_{3}\left(I_{3}-I_{1}\right) / I_{1}
\end{equation}