Torque-free rotation — Euler's equations, asymmetric top: Exercises

Quick toolbox we will lean on the whole way down (every symbol is earned in the parent, restated here so you never have to leave):

Formula The three tools

Euler's equations (torque-free) — the equations of motion, one per body axis: $I_{1} \overset{ω}{˙}_{1} = (I_{2} - I_{3}) ω_{2} ω_{3}, I_{2} \overset{ω}{˙}_{2} = (I_{3} - I_{1}) ω_{3} ω_{1}, I_{3} \overset{ω}{˙}_{3} = (I_{1} - I_{2}) ω_{1} ω_{2}$ Here $I_{1}, I_{2}, I_{3}$ are the principal moments of inertia (three fixed numbers describing how the mass is spread about each body axis), and $ω_{1}, ω_{2}, ω_{3}$ are the components of the spin vector $ω$ along those axes. A dot means "rate of change in time". Two conserved scalars: $2 T = I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2} (energy), L^{2} = I_{1}^{2} ω_{1}^{2} + I_{2}^{2} ω_{2}^{2} + I_{3}^{2} ω_{3}^{2} (angular momentum squared)$ Stability sign for a spin nearly about axis $b$ (with $a, c$ the other two): $k = \frac{( I _{a} - I _{b} ) ( I _{b} - I _{c} )}{I _{a} I _{c}} Ω^{2} < 0 \Rightarrow stable (oscillate), > 0 \Rightarrow unstable (grow) .$

Related depth lives in Inertia tensor and principal axes, Stability analysis and linearization, and Symmetric top and gyroscopic precession.

Level 1 — Recognition

Problem 1.1

A rigid body has $I_{1} = I_{2} = I_{3} = 4 kg\cdotm^{2}$ (a uniform ball). It is spun with $ω = (3, - 1, 2) rad/s$ and released with no external torque. What is $\dot{ω}$ at that instant?

Recall Solution 1.1

WHAT to notice: all three moments are equal. Look at each Euler equation — every right-hand side carries a difference of moments like $(I_{2} - I_{3})$ . When all $I_{i}$ are equal, every such difference is zero. $I_{1} \overset{ω}{˙}_{1} = (I_{2} - I_{3}) ω_{2} ω_{3} = (4 - 4) ω_{2} ω_{3} = 0 \Rightarrow \overset{ω}{˙}_{1} = 0,$ and identically $\overset{ω}{˙}_{2} = \overset{ω}{˙}_{3} = 0$ . Answer: $\dot{ω} = (0, 0, 0)$ . The spin never changes — a sphere has no wobble because there is no moment-difference to couple the axes.

Problem 1.2

For a body with $I_{1} = 1, I_{2} = 2, I_{3} = 3$ , which single principal axis, if spun about, is unstable? Just name it.

Recall Solution 1.2

Order the moments: smallest is $I_{1}$ , largest is $I_{3}$ , intermediate is $I_{2}$ . The tennis-racket theorem says rotation about the intermediate axis is unstable. Answer: axis 2 (the $I_{2} = 2$ axis).

Level 2 — Application

Problem 2.1

A symmetric top has $I_{1} = I_{2} = I_{⊥} = 2 kg\cdotm^{2}$ and $I_{3} = 5 kg\cdotm^{2}$ . It spins with $ω_{3} = 10 rad/s$ . Find the body precession rate $Ω_{b}$ (the rate at which $ω_{1}, ω_{2}$ rotate around axis 3).

Recall Solution 2.1

WHY this works: with $I_{1} = I_{2}$ , Euler's third equation gives $I_{3} \overset{ω}{˙}_{3} = (I_{1} - I_{2}) ω_{1} ω_{2} = 0$ , so $ω_{3}$ is frozen. The first two equations then read $\overset{ω}{˙}_{1} = - Ω_{b} ω_{2}, \overset{ω}{˙}_{2} = + Ω_{b} ω_{1}, Ω_{b} = \frac{I _{3} - I _{⊥}}{I _{⊥}} ω_{3} .$ This is uniform circular motion of the pair $(ω_{1}, ω_{2})$ . $Ω_{b} = \frac{5 - 2}{2} \times 10 = \frac{3}{2} \times 10 = 15 rad/s .$ Answer: $Ω_{b} = 15 rad/s$ — $ω$ sweeps a cone about axis 3 fifteen radians per second.

Problem 2.2

Body with $I_{1} = 1, I_{2} = 2, I_{3} = 3$ at an instant has $ω = (2, 0, 3) rad/s$ . Compute $2 T$ and $L^{2}$ and state their values at all later times.

Recall Solution 2.2

Plug straight into the conserved scalars. $2 T = I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2} = 1 (2^{2}) + 2 (0^{2}) + 3 (3^{2}) = 4 + 0 + 27 = 31 J (i.e. 2 T) .$ $L^{2} = I_{1}^{2} ω_{1}^{2} + I_{2}^{2} ω_{2}^{2} + I_{3}^{2} ω_{3}^{2} = 1 (4) + 4 (0) + 9 (9) = 4 + 0 + 81 = 85.$ WHY constant: we proved $d T / d t = 0$ (the moment-differences telescope) and $L$ is fixed in space so $L^{2}$ cannot change. Both hold for all time. Answer: $2 T = 31 J$ , $L^{2} = 85 kg^{2} m^{4} / s^{2}$ , forever.

Level 3 — Analysis

Problem 3.1

For $I_{1} = 1, I_{2} = 2, I_{3} = 3$ , a body spins nearly about axis 1 with $Ω = 8 rad/s$ . Compute the stability constant $k$ and the oscillation frequency of small disturbances.

Recall Solution 3.1

Spinning about axis 1, the "other two" are $2, 3$ . The linearised equation is $\overset{ω}{¨}_{2} = k ω_{2}$ with $k = \frac{( I _{2} - I _{1} ) ( I _{1} - I _{3} )}{I _{2} I _{3}} Ω^{2} = \frac{( 2 - 1 ) ( 1 - 3 )}{2 \cdot 3} (8)^{2} = \frac{( 1 ) ( - 2 )}{6} \times 64 = - \frac{128}{6} = - 21.33.$ Because $k < 0$ , write $k = - ν^{2}$ so the solution is $cos (ν t)$ : it oscillates, hence stable. $ν = ∣ k ∣ = 21.33 = 4.62 rad/s .$ Answer: $k = - 21.33$ , small wobbles oscillate at $ν \approx 4.62 rad/s$ — axis 1 (smallest $I$ ) is stable. ✅

Problem 3.2

Same body, but spin nearly about axis 2 with $Ω = 8 rad/s$ . Compute $k$ and the exponential growth rate; how long until a disturbance grows by a factor $e$ ?

Recall Solution 3.2

Now the "other two" are $1, 3$ : $k = \frac{( I _{1} - I _{2} ) ( I _{2} - I _{3} )}{I _{1} I _{3}} Ω^{2} = \frac{( 1 - 2 ) ( 2 - 3 )}{1 \cdot 3} (64) = \frac{( - 1 ) ( - 1 )}{3} \times 64 = \frac{64}{3} = 21.33 > 0.$ Positive $k$ means the solution is $e^{k t}$ — exponential growth, unstable. The growth rate is $k = 21.33 = 4.62 s^{- 1}, t_{e} = \frac{1}{k} = 0.217 s .$ Answer: $k = + 21.33$ , e-folding time $\approx 0.22 s$ — axis 2 (intermediate $I$ ) is unstable. ❌ This is the Dzhanibekov flip.

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

Level 4 — Synthesis

Problem 4.1

A body with $I_{1} = 1, I_{2} = 2, I_{3} = 4$ starts at $ω = (0, 0, 5) rad/s$ (pure spin about axis 3). Show that if instead it starts at $ω = (ϵ, 0, 5)$ with tiny $ϵ$ , the trajectory of $ω$ stays on a thin closed loop, and find the wobble frequency about axis 3.

Recall Solution 4.1

Strategy: axis 3 is the largest $I$ here, so we expect stability (closed loop). Confirm by linearising: keep $ω_{3} \approx Ω = 5$ constant, $ω_{1}, ω_{2}$ small. $\overset{ω}{˙}_{1} = \frac{I _{2} - I _{3}}{I _{1}} Ω ω_{2}, \overset{ω}{˙}_{2} = \frac{I _{3} - I _{1}}{I _{2}} Ω ω_{1} .$ Differentiate the first, substitute the second: $\overset{ω}{¨}_{1} = \frac{( I _{2} - I _{3} ) ( I _{3} - I _{1} )}{I _{1} I _{2}} Ω^{2} ω_{1} = \frac{( 2 - 4 ) ( 4 - 1 )}{1 \cdot 2} (25) ω_{1} = \frac{( - 2 ) ( 3 )}{2} (25) ω_{1} = - 75 ω_{1} .$ $k = - 75 < 0$ ⇒ oscillation, so $ω$ traces a small ellipse (a polhode loop) around axis 3 — it never runs away, confirming a thin closed loop. Wobble frequency: $ω_{wobble} = ∣ k ∣ = 75 = 8.66 rad/s .$ Answer: stable closed loop, wobble frequency $\approx 8.66 rad/s$ . The two conserved surfaces $2 T$ and $L^{2}$ intersect in a tight closed polhode near axis 3.

Problem 4.2

For the symmetric top of Problem 2.1 ( $I_{⊥} = 2, I_{3} = 5, ω_{3} = 10$ ), the transverse spin has amplitude $ω_{⊥} = ω_{1}^{2} + ω_{2}^{2} = 1 rad/s$ . Find (a) the half-angle of the cone $ω$ sweeps about axis 3 in the body frame, and (b) the conserved $L^{2}$ .

Recall Solution 4.2

(a) The spin vector makes angle $α$ with axis 3 where $tan α = \frac{ω _{⊥}}{ω _{3}} = \frac{1}{10} = 0.1 \Rightarrow α = arctan (0.1) = 5.7 1^{\circ} .$ Why $tan$ ? On the right triangle whose legs are $ω_{⊥}$ (side opposite the tilt) and $ω_{3}$ (side adjacent, along the axis), tangent = opposite/adjacent is exactly the ratio that encodes how far $ω$ tips from the axis. (b) With $ω_{1}^{2} + ω_{2}^{2} = 1$ and $I_{1} = I_{2} = 2$ : $L^{2} = I_{⊥}^{2} (ω_{1}^{2} + ω_{2}^{2}) + I_{3}^{2} ω_{3}^{2} = 2^{2} (1) + 5^{2} (100) = 4 + 2500 = 2504.$ Answer: (a) $α \approx 5.7 1^{\circ}$ ; (b) $L^{2} = 2504 kg^{2} m^{4} / s^{2}$ .

Level 5 — Mastery

Problem 5.1

A body has $I_{1} = 1, I_{2} = 2, I_{3} = 3$ . It is given $2 T = 31$ and $L^{2} = 85$ (the state from Problem 2.2). The separatrix — the borderline polhode that passes through the unstable axis 2 — occurs when $L^{2} = 2 T I_{2}$ . Check whether this exact state sits on the separatrix, above it (energy ellipsoid encloses axis 3), or below it (encloses axis 1), and interpret physically.

Recall Solution 5.1

Strategy — the ratio $L^{2} / (2 T)$ acts like an "effective $I$ ". Compute it and compare to the moments. $\frac{L ^{2}}{2 T} = \frac{85}{31} = 2.742.$ This lies between $I_{2} = 2$ and $I_{3} = 3$ . Interpretation: because $L^{2} / (2 T) > I_{2}$ , the intersection of the two ellipsoids is a polhode that loops around axis 3 (the largest- $I$ axis), not around axis 2. So this state is on the stable side, wrapping the large axis. Separatrix would require $L^{2} / (2 T) = I_{2} = 2$ exactly, i.e. $L^{2} = 2 T \cdot I_{2} = 31 \times 2 = 62$ . We have $85 \neq = 62$ , so we are not on the separatrix; we are safely above it (axis-3 loop). Answer: $L^{2} / (2 T) = 2.742$ lies between $I_{2}$ and $I_{3}$ ⇒ polhode encircles axis 3, stable side; separatrix value would be $L^{2} = 62$ .

Problem 5.2

Design check (Dzhanibekov timer). A wingnut in space is modelled as $I_{1} = 1, I_{2} = 2, I_{3} = 3 kg\cdotm^{2}$ , tossed spinning about the unstable axis 2 at $Ω = 12 rad/s$ with an unavoidable seed disturbance $ω_{1} (0) = 0.01 rad/s$ . Estimate how long until the disturbance reaches $ω_{1} = 1 rad/s$ (roughly "flip-scale"). Use the linear growth $ω_{1} (t) = ω_{1} (0) e^{k t}$ .

Recall Solution 5.2

Step 1 — growth rate. For axis 2: $k = \frac{( I _{1} - I _{2} ) ( I _{2} - I _{3} )}{I _{1} I _{3}} Ω^{2} = \frac{( - 1 ) ( - 1 )}{1 \cdot 3} (144) = \frac{144}{3} = 48, k = 48 = 6.928 s^{- 1} .$ Step 2 — invert the exponential. We need $ω_{1} (t) / ω_{1} (0) = 1/0.01 = 100$ : $e^{k t} = 100 \Rightarrow t = \frac{l n 100}{k} = \frac{4.6052}{6.928} = 0.6647 s .$ Why the log? The exponential growth $e^{k t}$ answers "how big after time $t$ "; taking the natural log inverts it — it answers the reverse question "how long to reach a given size". That is exactly what $ln$ undoes. Answer: roughly $0.66 s$ from toss to visible flip — the wingnut tumbles within a second, matching the famous ISS footage.

Recall Self-test index (close the page, answer from memory)

Which axis is unstable and why ::: The intermediate- $I$ axis, because the product $(I_{a} - I_{b}) (I_{b} - I_{c})$ is positive there, giving $k > 0$ and exponential growth. What two scalars are conserved in torque-free motion ::: $2 T = \sum I_{i} ω_{i}^{2}$ and $L^{2} = \sum I_{i}^{2} ω_{i}^{2}$ . Formula for symmetric-top body precession rate ::: $Ω_{b} = \frac{I _{3} - I _{⊥}}{I _{⊥}} ω_{3}$ . Why is $ω$ not constant even with zero torque ::: Because $L = I ω$ is conserved, not $ω$ ; a lopsided (non-sphere) $I$ makes constant $L$ force a moving $ω$ .