Gyroscope — steady precession derivation: Worked examples

This page is the drill hall. Before any formula appears, let me name every letter so a reader who arrived on THIS page from nowhere can follow from line one.

Definition The cast of symbols (every one, in words — defined BEFORE first use)

$m$ = mass of the top, in kilograms. How heavy it is.
$g$ = gravitational acceleration $= 9.8 m/s^{2}$ . How hard Earth pulls.
$ℓ$ = distance from the pivot point $O$ to the center of mass, along the axis, in metres. The "lever length."
$θ$ = tilt angle of the axis away from straight-up (the vertical). $θ = 0$ means bolt upright; $θ = 9 0^{\circ}$ means lying flat. See Euler Angles.
$I_{3}$ = axial moment of inertia: how hard it is to spin the top about its OWN axis. See Moment of Inertia Tensor.
$I_{1}$ = transverse moment of inertia: how hard it is to tip the whole top over sideways.
$ω_{s}$ = spin rate about the symmetry axis, in radians per second. How fast it whirls.
$Ω = \dot{ϕ}$ = precession rate: how fast the axis walks around the vertical circle. This is what we solve for.

Now that every symbol has a plain-word meaning and a picture (the tilted axis, the lever, the walk-around), here are the two tools the parent note built:

The picture behind both formulas — the tilted axis, gravity pulling at the center of mass, the resulting torque, and the sideways "walk" of the axis — is here. Keep referring back to it; every example below lives on this one diagram.

Figure — Gyroscope — steady precession derivation

Read it left to right: the blue arrow is the spin angular momentum $L_{s} = I_{3} ω_{s}$ along the leaning axis; the yellow dot is the center of mass a distance $ℓ$ out; the red arrow is gravity $m g$ pulling straight down; the green circle at the tip is the precession walk at rate $Ω$ ; the white arc is the tilt angle $θ$ .

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The nine worked examples below are each tagged with the cell(s) they cover, so together they blanket the whole table.

Cell	Case class	What makes it tricky
A	Standard fast top	plug straight into $Ω = m g ℓ / I_{3} ω_{s}$
B	Sign / direction of $Ω$	which way does it walk — clockwise or counter?
C	Limiting input: $ω_{s} \to \infty$	precession $\to 0$ (super-stubborn)
D	Limiting input: $ω_{s} \to$ small	discriminant goes negative — no steady precession
E	Degenerate tilt $θ = 9 0^{\circ}$	axis horizontal; does the formula survive?
F	Degenerate tilt $θ = 0^{\circ}$ (sleeping top)	axis vertical; $sin θ = 0$ everywhere
G	Exact vs approximate — both roots	fast & slow precession compared
H	Real-world word problem	translate English → symbols, keep units
I	Exam twist: solve for a hidden variable	given $Ω$ , find $ω_{s}$ or $I_{3}$

Worked examples

Example 1 — Cell A (standard fast top)

Worked example The bread-and-butter case

A gyroscope has $I_{3} = 0.10 kg\cdotm^{2}$ , spins at $ω_{s} = 40 rad/s$ , hangs $ℓ = 0.20$ m from the pivot, mass $m = 2.0$ kg. Find the precession rate $Ω$ .

Forecast: Guess — will $Ω$ be a big number (fast walk) or a small one (slow walk)? A fast spinner is stubborn, so expect $Ω$ small.

Step 1. Identify which formula. The spin is large, so use fast-top. Why this step? Fast-top is valid when $(I_{3} ω_{s})^{2} ≫ 4 I_{1} m g ℓ cos θ$ . With big $ω_{s}$ the approximation is safe and far simpler.

Step 2. Compute the numerator $m g ℓ = 2.0 \cdot 9.8 \cdot 0.20 = 3.92$ . Why this step? $m g ℓ$ IS the gravity torque's magnitude at $θ = 9 0^{\circ}$ — the "toppling effort" (the red arrow's turning power in the figure).

Step 3. Compute the denominator $I_{3} ω_{s} = 0.10 \cdot 40 = 4.0$ . Why this step? $I_{3} ω_{s}$ is the spin angular momentum $L_{s}$ — the blue arrow's length, the "stubbornness."

Step 4. Divide: $Ω = 3.92/4.0 = 0.98 rad/s$ .

Verify: Units: $\frac{kg \cdot ( m/s ^{2} ) \cdot m}{kg\cdotm ^{2} \cdot ( 1/ s )} = \frac{kg\cdotm ^{2} / s ^{2}}{kg\cdotm ^{2} / s} = \frac{1}{s}$ ✓. That's rad/s. Small number as forecast; period $2 π /0.98 \approx 6.4$ s — a leisurely sweep. ✓

Example 2 — Cell B (direction/sign)

Worked example Which way does it walk?

Same gyro as Example 1. The spin angular momentum $L_{s}$ points outward along the tilted axis (right-hand rule from the spin). The axis leans in the $+ x$ direction. Gravity torque is $τ = r \times m g$ . Which way (clockwise or counter-clockwise, viewed from above) does the axis precess, and what is $Ω$ 's sign?

Forecast: Torque is horizontal and perpendicular to the axis. Guess: the tip moves in the direction of $τ$ .

Step 1. Place $r = ℓ \overset{e}{^}_{3}$ leaning toward $+ x$ , so its horizontal part is $+ x$ , vertical part $+ z$ . Gravity $m g = - m g \overset{z}{^}$ . Why this step? We need actual vectors to take a cross product; a picture alone can mislead on sign. Figure s02 draws these three vectors explicitly.

Step 2. $τ = r \times m g$ . The horizontal $+ x$ piece crossed with $- \overset{z}{^}$ gives $\overset{x}{^} \times (- \overset{z}{^}) = + \overset{y}{^}$ . Why this step? Right-hand rule: $\overset{x}{^} \times \overset{z}{^} = - \overset{y}{^}$ , so $\overset{x}{^} \times (- \overset{z}{^}) = + \overset{y}{^}$ . Torque points $+ y$ — the green arrow in figure s02.

Step 3. Since $d L = τ d t$ , the horizontal part of $L$ (which points $+ x$ ) gets nudged toward $+ y$ . So the axis rotates from $+ x$ toward $+ y$ : counter-clockwise seen from above, $Ω > 0$ about $+ \overset{z}{^}$ . Why this step? Precession is nothing but $L$ 's tip chasing $τ$ . Direction of $Ω$ = direction from $L_{h}$ toward $τ$ .

Verify: Consistency check with $τ = Ω \times L$ : need $Ω \times L_{h}$ to equal $+ \overset{y}{^}$ . Take $Ω = Ω \overset{z}{^}$ , $L_{h} = L_{h} \overset{x}{^}$ : $\overset{z}{^} \times \overset{x}{^} = + \overset{y}{^}$ ✓. Magnitude $Ω = m g ℓ / (I_{3} ω_{s}) = 0.98$ rad/s as before. ✓

Example 3 — Cell C (limit $ω_{s} \to \infty$ )

Worked example The infinitely stubborn top

Take Example 1 but let the spin grow: $ω_{s} = 400$ , then $4000 rad/s$ . What happens to $Ω$ ?

Forecast: More spin = more stubborn = ??? Guess the direction of change.

Step 1. $Ω = m g ℓ / (I_{3} ω_{s}) = 3.92/ (0.10 \cdot 400) = 0.098 rad/s$ . Why this step? Same fast-top formula; only $ω_{s}$ changed.

Step 2. At $ω_{s} = 4000$ : $Ω = 3.92/ (0.10 \cdot 4000) = 0.0098 rad/s$ . Why this step? $Ω \propto 1/ ω_{s}$ , so ten-fold spin gives one-tenth precession.

Step 3. Limit: as $ω_{s} \to \infty$ , $Ω \to 0$ . Why this step? An infinitely spun top essentially freezes its axis in place — the ideal "gyroscopic compass" that points a fixed direction forever.

Verify: $Ω \cdot ω_{s} = m g ℓ / I_{3} = 39.2$ is constant regardless of $ω_{s}$ : $0.098 \cdot 400 = 39.2$ ✓, $0.0098 \cdot 4000 = 39.2$ ✓. The product is invariant. ✓

Example 4 — Cell D (limit small spin → no steady precession)

Worked example Spin too slow: reality check

Thin top: $I_{1} = 0.012$ , $I_{3} = 0.006 kg\cdotm^{2}$ , $m = 0.5$ kg, $ℓ = 0.05$ m, $θ = 3 0^{\circ}$ . Find the minimum spin below which steady precession is impossible.

Forecast: There should be a threshold — below it the top can't precess steadily, it must nutate or fall. Guess: tens of rad/s.

Step 1. Demand the discriminant of the exact quadratic be $\geq 0$ : $(I_{3} ω_{s})^{2} \geq 4 I_{1} m g ℓ cos θ$ . Why this step? $Ω_{\pm}$ are real only if the square root's contents are non-negative. Negative discriminant = no real precession rate = no steady motion. See Nutation.

Step 2. Compute the right side: $4 \cdot 0.012 \cdot 0.5 \cdot 9.8 \cdot 0.05 \cdot cos 3 0^{\circ} = 4 \cdot 0.012 \cdot 0.5 \cdot 9.8 \cdot 0.05 \cdot 0.8660 = 0.010184$ . Why this step? We plug all knowns; $cos 3 0^{\circ} = 0.8660$ .

Step 3. So $I_{3} ω_{s} \geq 0.010184 = 0.10092$ , giving $ω_{s} \geq 0.10092/0.006 = 16.8 rad/s$ . Why this step? Divide by $I_{3}$ to isolate $ω_{s}$ .

Verify: Units: numerator $kg\cdotm^{2} \cdot kg \cdot (m/s^{2}) \cdot m = (kg\cdotm^{2} / s)^{2} = kg\cdotm^{2} / s$ ; divide by $I_{3}$ (kg·m²) → 1/s ✓. Numeric $16.8$ rad/s, in the tens as forecast. ✓

Example 5 — Cell E (degenerate tilt $θ = 9 0^{\circ}$ )

Worked example Axis lying flat

The bicycle wheel of Example 1 is held with its axle horizontal, $θ = 9 0^{\circ}$ . Find $Ω$ from fast-top, and check the exact quadratic doesn't blow up.

Forecast: At $θ = 9 0^{\circ}$ the torque is maximal ( $m g ℓ sin 9 0^{\circ} = m g ℓ$ ). Does $Ω$ blow up too? Recall the $sin θ$ cancellation.

Step 1. Fast-top: $Ω = m g ℓ / (I_{3} ω_{s}) = 0.98$ rad/s — unchanged from Example 1. Why this step? The $sin θ$ from torque and the $sin θ$ from horizontal $L$ cancelled in the parent derivation, so $Ω$ is angle-free in the approximation.

Step 2. Exact quadratic: the $cos θ$ coefficient becomes $cos 9 0^{\circ} = 0$ . The quadratic degenerates to a linear equation: $- I_{3} ω_{s} Ω + m g ℓ = 0$ . Why this step? When the leading $Ω^{2}$ coefficient vanishes, "quadratic" collapses to linear — a case you must handle separately or the $/ cos θ$ divides by zero.

Step 3. Solve the linear form: $Ω = m g ℓ / (I_{3} ω_{s}) = 0.98$ rad/s — the exact and fast-top answers coincide exactly at $θ = 9 0^{\circ}$ . Why this step? With $cos θ = 0$ the $I_{1}$ term drops entirely, so there's no correction; the two formulas literally agree.

Verify: Plug $Ω = 0.98$ , $θ = 9 0^{\circ}$ into full quadratic: $I_{1} (0.98)^{2} (0) - (4.0) (0.98) + 3.92 = 0 - 3.92 + 3.92 = 0$ ✓. Exact. ✓

Example 6 — Cell F (degenerate tilt $θ = 0^{\circ}$ , sleeping top)

Worked example The upright "sleeping" top

A top spins perfectly upright, $θ = 0^{\circ}$ . What does the theory say about $Ω$ ?

Forecast: Upright means no lean, no horizontal offset, no toppling torque. Guess: precession is undefined or zero.

Step 1. Torque magnitude $m g ℓ sin θ = m g ℓ sin 0^{\circ} = 0$ . Why this step? The center of mass sits directly above the pivot, so gravity has zero lever arm — no torque to drive precession.

Step 2. In the fast-top equation $m g ℓ sin θ = Ω I_{3} ω_{s} sin θ$ , every term carries $sin θ = 0$ . The equation reads $0 = 0$ — $Ω$ is indeterminate, not forced to any value. Why this step? You cannot cancel $sin θ$ when it is zero. The "steady precession" question becomes vacuous: any $Ω$ (including $0$ ) satisfies the balance trivially.

Step 3. Physically the top just spins in place ("sleeping"). Stability is a separate question decided by Nutation: the upright is stable if $(I_{3} ω_{s})^{2} > 4 I_{1} m g ℓ$ (the $θ = 0$ discriminant with $cos 0 = 1$ ). Why this step? Whether a slightly-tipped sleeping top returns upright or falls needs the discriminant at $θ = 0$ , our threshold from Cell D generalised.

Verify: Discriminant at $θ = 0$ for Example-4 top: $4 I_{1} m g ℓ \cdot 1 = 4 \cdot 0.012 \cdot 0.5 \cdot 9.8 \cdot 0.05 = 0.01176$ ; need $(I_{3} ω_{s})^{2} > 0.01176$ , i.e. $ω_{s} > 0.01176 /0.006 = 18.1$ rad/s. Slightly higher threshold than the $θ = 3 0^{\circ}$ case (16.8), because $cos 0 > cos 3 0^{\circ}$ ✓.

Example 7 — Cell G (exact: both roots, fast & slow)

Worked example Two precession rates at once

Top with $I_{1} = 0.012$ , $I_{3} = 0.006 kg\cdotm^{2}$ , $m = 0.5$ kg, $ℓ = 0.05$ m, $θ = 3 0^{\circ}$ , spun at $ω_{s} = 50 rad/s$ (well above the 16.8 threshold). Find both exact precession rates $Ω_{+}$ and $Ω_{-}$ .

Forecast: One root should be the familiar slow precession; the other a fast one. Guess which sign of the $\pm$ gives the slow one.

Step 1. Compute $I_{3} ω_{s} = 0.006 \cdot 50 = 0.30$ . Why this step? This is the shared numerator term and the leading term inside the root.

Step 2. Discriminant $(I_{3} ω_{s})^{2} - 4 I_{1} m g ℓ cos θ = 0.3 0^{2} - 0.010184 = 0.09 - 0.010184 = 0.079816$ ; its root is $0.28252$ . Why this step? We reuse the same $4 I_{1} m g ℓ cos θ = 0.010184$ from Example 4 (same top, same angle).

Step 3. Denominator $2 I_{1} cos θ = 2 \cdot 0.012 \cdot 0.8660 = 0.020785$ . Why this step? This scales both roots.

Step 4. $Ω_{-} = (0.30 - 0.28252) /0.020785 = 0.01748/0.020785 = 0.841 rad/s$ (slow). $Ω_{+} = (0.30 + 0.28252) /0.020785 = 0.58252/0.020785 = 28.03 rad/s$ (fast). Why this step? The $-$ root, from a near-cancellation in the numerator, gives the small (physically usual) precession; $+$ gives the rarely-seen fast branch.

Verify: Fast-top prediction for $Ω_{-}$ : $m g ℓ / (I_{3} ω_{s}) = (0.5 \cdot 9.8 \cdot 0.05) /0.30 = 0.245/0.30 = 0.817$ rad/s — within $3%$ of the exact $0.841$ ✓. Both roots plugged into the quadratic $I_{1} Ω^{2} cos θ - I_{3} ω_{s} Ω + m g ℓ$ give $\approx 0$ . See Symmetric Top and Lagrangian Mechanics. ✓

Example 8 — Cell H (real-world word problem)

Worked example Ship's gyrocompass

A gyrocompass rotor is a solid disk, mass $m = 6.0$ kg, radius $R = 0.15$ m, spun at $ω_{s} = 300 rad/s$ . Its center of mass is offset $ℓ = 0.010$ m from the pivot by a small pendulous weight (this is what makes it seek north). Find its precession rate.

Forecast: Fast spin + tiny offset → expect a very slow precession, seconds-to-minutes per turn.

Step 1. Axial moment of a solid disk about its axis: $I_{3} = \frac{1}{2} m R^{2} = \frac{1}{2} \cdot 6.0 \cdot 0.1 5^{2} = 0.0675 kg\cdotm^{2}$ . Why this step? We translate "solid disk" into its moment of inertia; the axial value uses the $\frac{1}{2} m R^{2}$ standard result.

Step 2. Numerator $m g ℓ = 6.0 \cdot 9.8 \cdot 0.010 = 0.588$ . Why this step? The pendulous offset $ℓ$ is the lever arm producing the aligning torque.

Step 3. $Ω = m g ℓ / (I_{3} ω_{s}) = 0.588/ (0.0675 \cdot 300) = 0.588/20.25 = 0.02904 rad/s$ . Why this step? Fast-top applies overwhelmingly here ( $ω_{s}$ huge, $ℓ$ tiny), so we divide the toppling torque by the spin angular momentum.

Verify: Units 1/s ✓. Period $2 π /0.02904 = 216$ s ≈ 3.6 minutes per revolution — slow, as forecast, which is exactly why gyrocompasses give a steady heading. ✓

Example 9 — Cell I (exam twist: solve for a hidden variable)

Worked example Back-solve for spin

An exam gives you the observed precession: a gyro precesses with period $T = 4.0$ s. Its data: $I_{3} = 0.050 kg\cdotm^{2}$ , $m = 1.5$ kg, $ℓ = 0.12$ m. Find the spin $ω_{s}$ it must have.

Forecast: Rearrange $Ω = m g ℓ / (I_{3} ω_{s})$ to isolate $ω_{s}$ . Expect a few tens of rad/s.

Step 1. Convert period to $Ω$ : $Ω = 2 π / T = 2 π /4.0 = 1.5708 rad/s$ . Why this step? The formula uses angular rate $Ω$ , not period; $Ω = 2 π / T$ by definition of one revolution.

Step 2. Rearrange: $ω_{s} = m g ℓ / (I_{3} Ω)$ . Why this step? Algebra — multiply both sides by $ω_{s}$ and divide by $I_{3} Ω$ to solve for the unknown.

Step 3. Numerator $m g ℓ = 1.5 \cdot 9.8 \cdot 0.12 = 1.764$ . Then $ω_{s} = 1.764/ (0.050 \cdot 1.5708) = 1.764/0.078540 = 22.46 rad/s$ . Why this step? Straight substitution once rearranged.

Verify: Plug back: $Ω = m g ℓ / (I_{3} ω_{s}) = 1.764/ (0.050 \cdot 22.46) = 1.764/1.123 = 1.571$ rad/s → period $2 π /1.571 = 4.0$ s ✓. Round-trip closes. ✓

Recall Quick self-test (cover the answers)

Which formula for a fast top? ::: $Ω = m g ℓ / (I_{3} ω_{s})$ Does $Ω$ depend on tilt $θ$ in the fast-top limit? ::: No — the $sin θ$ cancels. What happens to $Ω$ as $ω_{s} \to \infty$ ? ::: $Ω \to 0$ (axis freezes). Condition for steady precession to exist? ::: $(I_{3} ω_{s})^{2} \geq 4 I_{1} m g ℓ cos θ$ At $θ = 9 0^{\circ}$ , how do exact and fast-top compare? ::: They agree exactly ( $cos 9 0^{\circ} = 0$ ). At $θ = 0^{\circ}$ , what is the torque? ::: Zero — precession is indeterminate (sleeping top).