Question bank — Torque-free rotation — Euler's equations, asymmetric top

False. What is conserved in space is $\vec{L}$, not $\vec{\omega }$; since $\vec{L}=\mathbf{I}\vec{\omega }$ and the body's orientation changes $\mathbf{I}$ in the lab, $\vec{\omega }$ generally moves even while $\vec{L}$ is frozen.

False. $\vec{L}$ is constant in the space frame; in the body frame it moves because the body's axes rotate underneath it — that is exactly what $\vec{\omega }\times \vec{L}$ in Euler's equations tracks.

False. They are two different ellipsoids in $\omega$-space (semi-axes scale as $1/\sqrt{I_i}$ versus $1/I_i$); the real motion lives on their intersection, the polhode.

False. Every Euler right-hand side has a factor like $(I_2-I_3)$; with all moments equal these vanish, so $\dot{\vec{\omega }}=0$ and $\vec{\omega }$ is frozen in the body — no wobble.

True. With $\Omega$ the unperturbed spin rate, the linearized growth constant $k=\frac{(I_3-I_2)(I_2-I_1)}{I_1{I}_3}{\Omega }^2$ is a product of two positive factors, so $k>0$ gives exponential growth (the Dzhanibekov / tennis-racket effect).

True (in the diagonal form given). Off the principal axes $\mathbf{I}$ is not diagonal, $L_i\ne I_i{\omega }_i$, and messy off-diagonal terms appear; the clean cyclic form requires principal axes.

False as a causal claim. Both $T$ and $\vec{L}$ follow independently from $\vec{N}=0$; energy conservation is proved directly by substituting Euler's equations into $\dot{T}$, without ever invoking $\vec{L}$ conservation.

True. Euler-3 reads $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2=0$ when $I_1=I_2$, so ${\omega }_3$ is frozen and the transverse part precesses around it.

False. It is a linear instability: any nonzero perturbation, however tiny, grows exponentially like $e^{\sqrt{k}t}$ until it dominates.

The error jumps from $\dot{\vec{L}}=0$ to $\dot{\vec{\omega }}=0$. Those are equal only when $\mathbf{I}$ is constant and $\vec{L}\parallel \vec{\omega }$; for a non-spherical body neither holds, so $\vec{\omega }$ moves.

The right-hand side terms $(I_i-I_j){\omega }_j{\omega }_k$ are internal geometric coupling from the $\vec{\omega }\times \vec{L}$ transport term, not external work. With $\vec{N}=0$ the system is still torque-free.

Cyclic symmetry moves the labels, but stability depends on the sign of $(I_a-I_b)(I_b-I_c)$, which flips only for the intermediate axis. Symmetry of form ≠ symmetry of behaviour.

Changing direction at fixed magnitudes costs no energy. Concretely $\dot{T}=\sum I_i{\omega }_i{\dot{\omega }}_i$; substituting Euler's equations gives ${\omega }_1{\omega }_2{\omega }_3[(I_2-I_3)+(I_3-I_1)+(I_1-I_2)]$, and the bracket cancels term-by-term — $+I_2-I_3+I_3-I_1+I_1-I_2=0$ — so $\dot{T}=0$ exactly.

No — free precession of a symmetric top is torque-free; with transverse moment $I_{\perp }\equiv I_1=I_2$, the spin ${\omega }_3$ stays constant and the transverse spin circles at the body rate ${\Omega }_b=\frac{I_3-I_{\perp }}{I_{\perp }}{\omega }_3$, driven entirely by Earth's own asymmetry.

Backwards — axis-3 spin means ${\omega }_3\approx \Omega$ (the unperturbed spin rate) is the large term we hold constant, while ${\omega }_1,{\omega }_2$ are the small perturbations we track.

Because in the body frame the inertia tensor is a constant diagonal $\mathrm{diag}(I_1,I_2,I_3)$; in the lab it changes every instant as the body reorients, which would make $\dot{\vec{L}}=\mathbf{I}\dot{\vec{\omega }}+\dot{\mathbf{I}}\vec{\omega }$ intractable.

A vector can change for two reasons: it truly changes (body-frame derivative) or our rotating axes sweep past a fixed vector, making it appear to change; $\vec{\omega }\times \vec{A}$ is that apparent change. See Angular momentum in rotating frames.

The coupling comes from $\vec{\omega }\times \vec{L}$, a cross product of two vectors each linear in $\omega$; the result is quadratic, so each equation is fed by a product of the other two components.

On the intersection of the two ellipsoids, the polhodes near both the max and min inertia axes are small closed loops, so $\vec{\omega }$ orbits nearby — a bounded, oscillatory (stable) motion, not runaway growth. See Stability analysis and linearization.

Spinning about the intermediate-inertia axis gives $k>0$, so infinitesimal wobble grows exponentially; the phone rapidly swaps orientation — the Dzhanibekov effect in your hand.

Because near axis-3 spin the two small equations are linear in ${\omega }_1,{\omega }_2$: ${\dot{\omega }}_1=a{\omega }_2$ and ${\dot{\omega }}_2=b{\omega }_1$ with constant coefficients $a,b$ (constant because ${\omega }_3\approx \Omega$ is held fixed and ${\omega }_1{\omega }_2$, being second-order-small, is dropped). Differentiating ${\dot{\omega }}_1=a{\omega }_2$ gives ${\ddot{\omega }}_1=a{\dot{\omega }}_2=ab{\omega }_1$, so ${\omega }_2$ is eliminated and $k=ab$. The elimination works only because the coefficients are constants — it relies on $\Omega$ being (to first order) fixed.

$\vec{L}=\mathbf{I}\vec{\omega }$ is parallel to $\vec{\omega }$ only when $\vec{\omega }$ points along a principal axis (an eigenvector of $\mathbf{I}$); a generic $\vec{\omega }$ mixes axes with different $I_i$, tilting $\vec{L}$ away.

Every right-hand side vanishes, so $\dot{\vec{\omega }}=0$: the spin arrow is frozen in the body and (since $\vec{L}\parallel \vec{\omega }$ always) also frozen in space.

It stays there forever — a principal axis is a fixed point of Euler's equations (all products ${\omega }_j{\omega }_k$ that would kick it are zero), stable or unstable depending on the axis.

It degenerates into two curves that cross at the unstable axis; a body launched exactly on it would asymptotically approach the middle axis but any deviation sends it around, marking the boundary between the two stable families.

The growth constant $k\propto (I_2-I_1)\to 0$, so the instability weakens and vanishes; at exact symmetry there is no unique "intermediate" axis and the motion becomes steady free precession.

Then $k\propto {\Omega }^2=0$: no coupling, no wobble, no instability — a body at rest stays at rest, consistent with $\dot{\vec{\omega }}=0$.

The body precession rate ${\Omega }_b\propto {\omega }_3=0$, so ${\omega }_1,{\omega }_2$ are frozen: the body simply spins steadily about a fixed transverse axis with no precession. See Symmetric top and gyroscopic precession.

Rotational invariance of the system — by Noether's theorem, symmetry under rotations of space guarantees conservation of angular momentum $\vec{L}$.