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Free Precession

Using Conservation Laws and Modeling to Understand Unexpected Dynamics

The case of the spinning T-handle in zero-g
  • The unexpected behavior of this T-handle is a striking example of free precession - a subject usually covered in upper division Classical Mechanics courses.
  • Attempts to explain the dynamics of a spinning rigid body using $F=ma$ such as Tao's are not very satisfactory (see the 2019 update in Tao's post).
  • A simple and more correct explanation using conservation of energy and angular momentum leads to a quantitative model of the dynamics.

See physics educator Derek Muller's video, which is based on mathematician Terrence Tao's post.

Angular momentum and energy of a spinning body
ω r v dm
  • Velocity $\vb{v} = \vb*{\omega}\times\vb{r}$
  • Angular momentum $\vb{L} = \int dm\, \vb{r}\times\vb{v}$
  • Energy $E = \frac{1}{2} \int dm\, \vb{v}\cdot\vb{v} = \frac{1}{2} \vb*{\omega}\cdot\vb{L}$
  • Since $\vb{L}$ is a linear function of $\vb*{\omega}$, \[ \begin{pmatrix}L_x \\ L_y \\ L_z\end{pmatrix} = \begin{pmatrix}I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz}\end{pmatrix} \begin{pmatrix}\omega_x \\ \omega_y \\ \omega_z\end{pmatrix} \]
  • Working out the $I_{ij}$ from the definition of $\vb{L}$ you find the moment of inertia matrix is symmetric: \[ I_{xx} = \int dm\, (y^2+z^2),\quad I_{xy} = I_{yx} = -\int dm\, x y,\quad ... \]
  • Since $E \geq 0$, $I_{jk}$ is a positive definite matrix
Spinning bodies have three principal axes
x y z ω ωx ωy ωz L Lx Ly Lz
  • $I_{ij}$ has three orthogonal directions for which the vectors $\vb{L}$ and $\vb*{\omega}$ are parallel (eigenvectors)
  • Assume the $x$, $y$, and $z$ axes are along these principal axes of the body, so \[ (L_x,L_y,L_z) = (I_x\omega_x,I_y\omega_y,I_z\omega_z) \] Also assume $I_x \leq I_y \leq I_z$
  • Since the body is spinning, these $(x, y, z)$ axes are rotating - but this is the only way to make the components of $I_{ij}$ constant
  • The total angular momentum $L = ||\vb{L}||$ is \[ L^2 = L_x^2 + L_y^2 + L_z^2 \]
  • The total kinetic energy $E$ is \[ 2E = L_x^2/I_x + L_y^2/I_y + L_z^2/I_z \]
Relating rotating and stationary vector coordinates
ω A A ω×
  • Imagine a stationary coordinate frame $(x',y',z')$ to go along with our rotating body frame $(x,y,z)$
  • Relative to these primed coordinates, the time rate of change of the coordinates of any vector are \[ \dot{\vb{A}}' = \dot{\vb{A}} + \vb*{\omega}\times\vb{A} \]

  • Warning! Don't the vectors $\vb{L}$, $\vb*{\omega}$, and $\vb{v}$ really belong to the stationary primed system? Hint: Yes, but at any single instant of time, we can put $(x',y',z')=(x,y,z)$...
  • Conservation of angular momentum is $\dot{\vb{L}}'=0$, or \[ \dot{\vb{L}} = -\vb*{\omega}\times\vb{L} \] That is, $\vb{L}$ in the spinning frame must change so that $\vb{L}'$ stays fixed.
Spin must change to conserve angular momentum
  • The components of $\dot{\vb{L}} = -\vb*{\omega}\times\vb{L}$ are the three equations \[ \dot{L}_i = (1/I_k - 1/I_j) L_j L_k = J_{kj} L_j L_k \] where $ijk$ are $xyz$ or cyclic permutations $yzx$ or $zxy$, and $J_{kj} \equiv 1/I_k - 1/I_j.$
  • Important checks: $\dot{(L^2)}=2\vb{L}\cdot\dot{\vb{L}}=0$
    Using components you can check that $\dot{E}=0$ as well.
  • In $(L_x,L_y,L_z)$ space, $L^2$ constant is a sphere and $E$ constant is an ellipsoid. The $\vb{L}$ vector can only move along their intersection.
  • The three projections into the $xy$, $yz$, and $zx$ planes are just conics: \[ \begin{aligned} J_{xz}L_x^2 + J_{yz}L_y^2 &= 2E - L^2/I_z \\ J_{xy}L_x^2 - J_{yz}L_z^2 &= 2E - L^2/I_y \\ J_{xy}L_y^2 + J_{xz}L_z^2 &= L^2/I_x - 2E \end{aligned} \]
Now the T-handle's motion makes sense...
  • In the rotating frame, the $\vb{L}$ vector traces some curve on the sphere, depending on the value of $E$.
  • When $\vb{L}$ starts very close to any of the principal axes, it changes very slowly since $\dot{\vb{L}}$ is near zero.
  • The projections of these paths into the $xy$ and $yz$ planes are ellipses, but the projection into the $zx$ plane is a hyperbola!
  • When $\vb{L}$ starts very near the $+y$ axis, its hyperbola eventually takes it to the $-y$ axis. It moves quickly from plus to minus or minus to plus, and slowly when close to either $y$ direction.
Iz= Ix= 0.75 1.25 z z x x y y
Here's one way to build a numerical model

Plan (many other possibilities!):

  1. Choose units so $I_y=||\vb{L}||=1$.
  2. If $2E \ge L^2/I_y$, choose $k$ to be $z$, otherwise $x$, so that \[ |J_{ik}|L_i^2 + |J_{jk}|L_j^2 = |2E - L^2/I_k| \]
  3. Step $(L_i, L_j)$ forward in time to second order: \[ \Delta L_i = (\dot{L_i} + \tfrac{1}{2}\ddot{L_i}\Delta t)\Delta t \]
  4. Renormalize $(L_i, L_j)$ back onto its ellipse.
  5. Update $L_k$ to keep $||\vb{L}||=1$.
E z z x x y y
Studying your model
  1. With the model, you can explore changing the various parameters: the moments of inertia $I_i$, and the value of $2E-L^2/I_y$.
  2. In addition to being positive, the moments of inertia of any real body must satisfy the triangle inequalities $I_i \le I_j+I_k$ for all permutations $ijk$ of $xyz$. Can you see why?
  3. No real body is perfectly rigid. When the angular velocity vector changes in body fixed (rotating) coordinates, changing stresses cause the body to flex slightly. This flexing generates a small amount of heat which slowly reduces $E$. Drag the $E$ slider slowly downward to watch what happens.
  4. Does this numerical model quantitatively explain the T-handle video? How much can you check with and without more detailed knowledge about the T-handle construction?
Math connects a surprising variety of phenomena

There is an elegant solution to the equations for $\vb{L}$ involving the Jacobi elliptic functions - doubly periodic analytic functions that generalize trig and exponential functions.

A rigid pendulum with an amplitude near 180 degrees turns out to be an exact analog of the spinning T-handle.

peirce quincuncial map

Finally, one of my favorite map projections - the Peirce quincuncial - is based on elliptic functions.