Exercises — Rigid body dynamics — Euler angles, Euler's equations of motion

Order: rotate by $\varphi$ about the space $Z$-axis, then by $\theta$ about the new $x$-axis (the line of nodes), then by $\psi$ about the final body $z$-axis. So $R=R_z(\psi )R_x(\theta )R_z(\varphi )$.

• $\varphi$ = precession: how the tilted top sweeps around the vertical, like a clock hand.
• $\theta$ = nutation: the tilt of the top's axis away from vertical.
• $\psi$ = spin: how fast the top turns on its own point.

Every coupling coefficient $(I_i-I_j)$ is zero, so all three equations collapse to $I_i{\dot{\omega }}_i=0$, i.e. ${\dot{\omega }}_1={\dot{\omega }}_2={\dot{\omega }}_3=0$. The angular velocity is constant about any axis — a free sphere spins forever about whatever axis you gave it. No wobble is possible.

Equation 3: $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2=0$ because $I_1=I_2$. Hence ${\omega }_3=\text{const}$ — the spin about the symmetry axis is frozen.

Use the 3-1-3 formulas. With $\psi =0$: $\sin \psi =0,\cos \psi =1$. With $\theta =60^{\circ }$: $\sin \theta =\frac{\sqrt{3}}{2}\approx 0.8660$, $\cos \theta =\frac{1}{2}$. ${\omega }_1=\dot{\varphi }\sin \theta \sin \psi +\dot{\theta }\cos \psi =4(0.8660)(0)+0=0.$ ${\omega }_2=\dot{\varphi }\sin \theta \cos \psi -\dot{\theta }\sin \psi =4(0.8660)(1)-0=3.464\text{rad/s}.$ ${\omega }_3=\dot{\varphi }\cos \theta +\dot{\psi }=4(0.5)+10=12\text{rad/s}.$ So $\mathbf{\omega }=(0,3.464,12)\text{rad/s}$.

$\Omega =\frac{3.0-2.0}{2.0}(6.0)=\frac{1.0}{2.0}(6.0)=3.0\text{rad/s}.$ $T=\frac{2\pi }{3.0}=2.094\text{s}.$ The transverse part $({\omega }_1,{\omega }_2)$ traces a circle about the body-3 axis once every $2.09$ s.

At $t=0$: ${\omega }_1(0)=A\cos 0=A=0.4\text{rad/s}$, and ${\omega }_2(0)=A\sin 0=0$ ✓. At $t=T/4$, $\Omega t=\Omega \cdot \frac{2\pi }{4\Omega }=\frac{\pi }{2}$, so ${\omega }_1=A\cos \frac{\pi }{2}=0$ and ${\omega }_2=A\sin \frac{\pi }{2}=A=0.4\text{rad/s}$. The tip of the transverse vector has rotated a quarter turn: from pointing along axis 1 to along axis 2, magnitude preserved. See the circle below.

With $I_1=I_2$ and $\tau =0$:

Torque-free: $I_1{\dot{\omega }}_1=(I_2-I_3){\omega }_2{\omega }_3$ and $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2$. Treat ${\omega }_2\approx \text{const}$ (large), ${\omega }_1,{\omega }_3$ small. Differentiate the first and substitute the third: $I_1{\ddot{\omega }}_1=(I_2-I_3){\omega }_2{\dot{\omega }}_3=(I_2-I_3){\omega }_2\cdot \frac{(I_1-I_2){\omega }_1{\omega }_2}{I_3}.$ ${\ddot{\omega }}_1=\frac{(I_2-I_3)(I_1-I_2)}{I_1{I}_3}{\omega }_2^2{\omega }_1\equiv k{\omega }_1.$ Signs: $I_2-I_3<0$ and $I_1-I_2<0$, product $>0$, and $\frac{{\omega }_2^2}{I_1{I}_3}>0$, so $k>0$. ${\ddot{\omega }}_1=+k{\omega }_1$ with $k>0$ has exponentially growing solutions $e^{\sqrt{k}t}$ → axis 2 is unstable. That's the tumbling middle-axis (Dzhanibekov) effect.

Spin about axis 1: ${\omega }_1\approx \text{const}$ large, ${\omega }_2,{\omega }_3$ small. $I_2{\dot{\omega }}_2=(I_3-I_1){\omega }_3{\omega }_1,I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2.$ Differentiate the first, sub the second: $I_2{\ddot{\omega }}_2=(I_3-I_1){\omega }_1{\dot{\omega }}_3=(I_3-I_1){\omega }_1\cdot \frac{(I_1-I_2){\omega }_1{\omega }_2}{I_3},$ ${\ddot{\omega }}_2=\frac{(I_3-I_1)(I_1-I_2)}{I_2{I}_3}{\omega }_1^2{\omega }_2.$ Signs: $I_3-I_1>0$, $I_1-I_2<0$, product $<0$. So ${\ddot{\omega }}_2=-c{\omega }_2$ with $c>0$ → oscillation at $\sqrt{c}$ → stable. Axis 3 (largest $I$) works out the same way (both factors again give a negative coefficient). Only the intermediate axis is unstable. See the diagram.

$\Omega =\frac{I_3-I_1}{I_1}{\omega }_3$, so the free-precession period is $T_{\Omega }=\frac{2\pi }{|\Omega |}=\frac{2\pi }{{\omega }_3}\cdot \frac{I_1}{|I_3-I_1|}=T_{\text{spin}}\cdot \frac{I_1}{|I_3-I_1|}.$ (a) Frisbee: $I_3-I_1=2I_1-I_1=I_1$, so $\frac{I_1}{|I_3-I_1|}=1$. Thus $T_{\Omega }=T_{\text{spin}}$ — it wobbles once per spin (a fast, obvious tumble). (b) Earth: $\frac{I_1}{I_3-I_1}=305$, so $T_{\Omega }=305\text{days}\approx 0.84\text{yr}$. This is the rigid-body Chandler prediction (~305 d); the real Chandler wobble is ~433 d because Earth is elastic/fluid, not perfectly rigid.

$2T=1(2^2)+2(1^2)+3(0^2)=4+2+0=6\Rightarrow T=3\text{J (per unit... using SI numbers)}.$ $L^2=1^2(2^2)+2^2(1^2)+3^2(0^2)=4+4+0=8\Rightarrow L=\sqrt{8}=\mathrm{2.828.}$ In $\mathbf{\omega }$-space, constant $2T$ is an ellipsoid and constant $L^2$ is another quadric; the true motion is confined to their intersection curve (the "polhode"). Because both are conserved with zero torque, $\mathbf{\omega }(t)$ can never leave that curve — this is the geometric backbone of the tennis-racket theorem.

${\dot{\omega }}_1=-\Omega {\omega }_2,{\dot{\omega }}_2=+\Omega {\omega }_1$ integrate to ${\omega }_1=A\cos \Omega t,{\omega }_2=A\sin \Omega t$. The tip advances from $+1$-axis toward $+2$-axis when $\Omega >0$ (counterclockwise, same sense the spin ${\omega }_3>0$ defines by the right-hand rule), and reverses when $\Omega <0$.

• Oblate ($I_3>I_1$): $\Omega =\frac{I_3-I_1}{I_1}{\omega }_3>0$ → same sense as spin.
• Prolate ($I_3<I_1$): $\Omega <0$ → opposite sense. This sign is observable: it distinguishes a tumbling frisbee (co-rotating wobble) from a tumbling pencil (counter-rotating wobble).

Physics: gravity torque about the pivot has magnitude $mg\ell \sin \theta$, perpendicular to the spin axis. For a fast top the angular momentum is dominated by the spin, $\mathbf{L}\approx I_3{\omega }_3\hat{z}$. Torque perpendicular to $\mathbf{L}$ can only rotate $\mathbf{L}$ sideways (not change its length) → steady precession. Sideways rate: the horizontal projection of $\mathbf{L}$ has length $I_3{\omega }_3\sin \theta$ and sweeps at rate $\dot{\varphi }$, so $|\dot{\mathbf{L}}|=\dot{\varphi }{I}_3{\omega }_3\sin \theta$. Set equal to the torque: $mg\ell \sin \theta =\dot{\varphi }{I}_3{\omega }_3\sin \theta \Rightarrow \boxed{\dot{\varphi }=\frac{mg\ell }{I_3{\omega }_3}}.$ The $\sin \theta$ cancels — steady precession rate is independent of tilt (for the fast-top approximation). Number: $\dot{\varphi }=\frac{0.20}{(4.0\times 10^{-3})(150)}=\frac{0.20}{0.60}=0.333\text{rad/s}.$ See Gyroscope precession for the full slow/fast branch analysis.

Along the motion, $2T=6$ and $L^2=8$ (from L4.2), and ${\omega }_1,{\omega }_2$ are real so ${\omega }_1^2\ge 0,{\omega }_2^2\ge 0$. Solve the two invariants for ${\omega }_1^2,{\omega }_2^2$ in terms of ${\omega }_3$: $I_1{\omega }_1^2+I_2{\omega }_2^2=2T-I_3{\omega }_3^2=6-3{\omega }_3^2,$ $I_1^2{\omega }_1^2+I_2^2{\omega }_2^2=L^2-I_3^2{\omega }_3^2=8-9{\omega }_3^2.$ With $I_1=1,I_2=2$: ${\omega }_1^2+2{\omega }_2^2=6-3{\omega }_3^2$ and ${\omega }_1^2+4{\omega }_2^2=8-9{\omega }_3^2$. Subtract: $2{\omega }_2^2=(8-9{\omega }_3^2)-(6-3{\omega }_3^2)=2-6{\omega }_3^2\Rightarrow {\omega }_2^2=1-3{\omega }_3^2.$ Then ${\omega }_1^2=(6-3{\omega }_3^2)-2{\omega }_2^2=6-3{\omega }_3^2-2(1-3{\omega }_3^2)=4+3{\omega }_3^2$. Require ${\omega }_2^2\ge 0$: $1-3{\omega }_3^2\ge 0\Rightarrow {\omega }_3^2\le \frac{1}{3}$, so ${\omega }_3^{\max }=\frac{1}{\sqrt{3}}=0.5774\text{rad/s}.$ (${\omega }_1^2=4+3{\omega }_3^2>0$ always, so it never limits.) The body's spin about axis 3 oscillates between $-0.577$ and $+0.577$ — the polhode is a closed loop, matching L4.2's picture.

Put ${\omega }_3=0$ into the L5.2 expressions: ${\omega }_2^2=1-0=1\Rightarrow {\omega }_2=1$ ✓ and ${\omega }_1^2=4+0=4\Rightarrow {\omega }_1=2$ ✓. Exactly ${\mathbf{\omega }}_0$. The initial point sits at the ${\omega }_3=0$ crossing of the polhode — the moment of maximum transverse spin, minimum axial spin. Self-consistent.

Free precession period in terms of spin period ::: $T_{\Omega }=T_{\text{spin}}\frac{I_1}{|I_3-I_1|}$ Fast-top precession rate ::: $\dot{\varphi }=\frac{mg\ell }{I_3{\omega }_3}$ (note $\sin \theta$ cancels) Which axis is unstable for $I_1<I_2<I_3$? ::: the intermediate axis, $I_2$