27.5: Steady Precession
- Page ID
- 30697
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Under what conditions will a top, spinning under gravity, precess at a steady rate? The constancy of \(L_{3}\), \(L_{Z}\) mean that \(\Omega_{3}=\dot{\phi} \cos \theta+\dot{\psi}\), and \(\Omega_{\mathrm{pr}}=\dot{\phi}\) are constants.
The \(\theta\) Lagrange equation is
\[I_{1}^{\prime} \ddot{\theta}=I_{1}^{\prime} \dot{\phi}^{2} \sin \theta \cos \theta-I_{3}(\dot{\phi} \cos \theta+\dot{\psi}) \dot{\phi} \sin \theta+M g \ell \sin \theta\]
For constant \(\theta\), \(\ddot{\theta}=0\), so, with \(\Omega_{3}=\dot{\phi} \cos \theta+\dot{\psi}\), and \(\Omega_{\mathrm{pr}}=\dot{\phi}\).
\[I_{1}^{\prime} \Omega_{p r}^{2} \cos \theta-I_{3} \Omega_{3} \Omega_{p r}+M g \ell=0 \label{eq3}\]
Since Equation \ref{eq3} is a quadratic equation for the precession rate, there are two solutions in general: on staring at a precessing top, this is a bit surprising! We know that for the top, when it’s precessing nicely, the spin rate \(\Omega_{3}\) far exceeds the precession rate \(\Omega_{p r}\). Assuming \(I_{1}^{\prime}, I_{3}\) to be of similar size, this means the first term in the quadratic is much smaller than the second. If we just drop the first term, we get the precession rate
\[\Omega_{\text {precess }(\mathrm{slow})}=\dfrac{M g \ell}{I_{3} \Omega_{3}}, \quad\left(\Omega_{3} \gg \Omega_{\text {precess }}\right)\]
Note that this is independent of angle—the torque varies as \(\sin \theta\), but so does the horizontal component of the angular momentum, which is what’s changing.
This is the familiar solution for a child’s fast-spinning top precessing slowly. But this is a quadratic equation, there’s another possibility: in this large \(\Omega_{3}\) limit, this other possibility is that \(\Omega_{p r} \text { is itself of order } \Omega_{3}\), so now in the equation the last term, the gravitational one, is negligible, and
\[\Omega_{\text {precess }(\text { fast })} \cong I_{3} \Omega_{3} / I_{1}^{\prime} \cos \theta\]
This is just the nutation of a free top! In fact, of course, both of these are approximate solutions, only exact in the limit of infinite spin (where one goes to zero, the other to infinity), and a more precise treatment will give corrections to each arising from the other. Landau indicates the leading order gravitational correction to the free body nutation mode.