Gyroscopic effect — precession of spinning top: Worked examples

This page is the drill hall for the parent topic. We take the one master formula and push it into every corner: signs, degenerate inputs, limits, a word problem, and an exam trap. Nothing here contradicts the parent — we just go wider.

Before any numbers: every symbol below was earned in the parent. If a word ( $L$ , torque, cross product) feels unfamiliar, open the linked note first — this page assumes you can read the formula as a sentence.

The scenario matrix

Every problem this topic can throw is one of these cells. Each worked example below is tagged with the cell it lands on, so together they cover the whole grid.

Cell	What varies	Where it bites
A. Standard positive	all quantities ordinary & positive	baseline number-crunch
B. Direction / sign	which way does it precess? (quadrant of $r \times g$ )	Cross Product and Right-Hand Rule
C. Tilt cancellation	change $θ$ , watch $Ω$ stay fixed	the $sin θ$ trap
C′. Over-tilt	$θ > 9 0^{\circ}$ (axis hangs below horizontal)	sign of the swing, not the rate
D. Limit: fast spin	$ω \to \infty$	$Ω \to 0$ (super-stable)
E. Degenerate: $θ = 0$	axis exactly vertical	$τ = 0$ , no precession
F. Degenerate: $ω = 0$	not spinning	formula blows up → it just falls
G. Word problem	real gyroscope, mixed units	navigation-style setup
H. Exam twist	scaling / ratio, no calculator	proportional reasoning

The worked examples

A + B — baseline number AND its direction

Worked example Example 1 — Cell A & B: the wheel that walks in a circle

A ring-like wheel: $m = 2 kg$ , radius $R = 0.25 m$ , so $I = m R^{2}$ . It spins at $ω = 30 rad/s$ and hangs from a cord with its centre of mass $r = 0.15 m$ from the pivot. Tilt $θ = 9 0^{\circ}$ (axle horizontal). Find $Ω$ and the direction of precession if $L$ points along $+ x$ and gravity is $- z$ .

Forecast: guess — will $Ω$ be bigger or smaller than $1 rad/s$ ? Which way does the free end swing?

$I = m R^{2} = 2 (0.25)^{2} = 0.125 kg\cdotm^{2}$ . Why this step? A ring puts all its mass at radius $R$ ; Moment of Inertia for a ring is $m R^{2}$ .
$L = I ω = 0.125 \times 30 = 3.75 kg\cdotm^{2} / s$ . Why this step? $L = I ω$ is the definition of spin angular momentum (Angular Momentum).
$τ = r m g = 0.15 \times 2 \times 9.8 = 2.94 N\cdotm$ . Why this step? Gravity acts at the centre of mass; torque about the pivot is (arm)×(force) with the axle horizontal so $sin θ = 1$ .
$Ω = \frac{τ}{L} = \frac{2.94}{3.75} = 0.784 rad/s$ . Why this step? Direct plug into the master formula.
Direction. $τ = r \times m g$ . With $r$ along $+ x$ and $g$ along $- z$ : $\overset{x}{^} \times (- \overset{z}{^}) = + \overset{y}{^}$ (since $\overset{x}{^} \times \overset{z}{^} = - \overset{y}{^}$ ). So $d L ∥ + y$ : the free tip swings from $+ x$ toward $+ y$ — counter-clockwise seen from above. Why this step? Direction of change of $L$ is always the direction of $τ$ ; never guess, compute the cross product (Cross Product and Right-Hand Rule, and see Vector nature of dL/dt).

Figure — Fig s01 (top view of the swing). This picture does what words cannot: it puts $L$ and $τ$ at a literal $9 0^{\circ}$ so you see that the push is sideways, not downward. Pivot at the origin; red arrow $L$ along $+ x$ ; mint arrow $τ$ (hence $d L$ ) toward $+ y$ ; butter-yellow arc = the tip curling counter-clockwise onto the dashed lavender circle it will trace. The pedagogy: the perpendicularity that makes precession counter-intuitive is spatial, so only a spatial figure settles it.

Alt-text: top-down view, red L-vector on +x axis, mint torque arrow on +y axis at 90°, yellow arc showing counter-clockwise swing onto a dashed lavender precession circle centred on the pivot.

Verify: $Ω < 1$ , matching the forecast that a decent spin gives lazy precession. Units: $\frac{N\cdotm}{kg\cdotm ^{2} / s} = \frac{kg\cdotm ^{2} / s ^{2}}{kg\cdotm ^{2} / s} = s^{- 1} = rad/s$ . ✓ And $\overset{x}{^} \times (- \overset{z}{^}) = + \overset{y}{^}$ confirms the swing.

C — the tilt cancellation trap

Worked example Example 2 — Cell C: change the tilt, keep the rate

Same wheel as Example 1 ( $r m g = 2.94 N\cdotm$ , $L = 3.75$ ). Now tilt it to $θ = 3 0^{\circ}$ . Compute $τ$ , the horizontal- $L$ radius, and $Ω$ . Compare to the $9 0^{\circ}$ case.

Forecast: the torque drops (less lever arm horizontally). Does $Ω$ drop too?

$τ = r m g sin θ = 2.94 \times sin 3 0^{\circ} = 2.94 \times 0.5 = 1.47 N\cdotm$ . Why this step? Only the component of gravity perpendicular to the axis makes torque; that brings in $sin θ$ .
Horizontal projection of $L$ has magnitude $L sin θ = 3.75 \times 0.5 = 1.875 kg\cdotm^{2} / s$ . Why this step? Precession is $L$ 's tip going around the vertical. Only the part of $L$ that lies in the horizontal plane actually circles; the vertical part just sits there. So the horizontal projection is literally the radius of the circle the tip travels on — see the figure, where the mint arrow $L sin θ$ is the flat shadow of the tilted red $L$ .
$Ω = \frac{τ}{L sin θ} = \frac{1.47}{1.875} = 0.784 rad/s$ . Why this step? The two $sin θ$ factors — one up top in $τ$ , one down below in the projection — cancel.

Figure — Fig s02 (why the tilt cancels). The figure earns its place by showing the two things that both scale with $sin θ$ in one frame, so the cancellation is visible rather than merely algebraic. Dotted line = vertical; red arrow = $L$ tilted by the lavender angle $θ = 3 0^{\circ}$ ; the flat mint arrow dropped onto the horizontal, of length $L sin θ$ , is the radius of the precession circle. Because the driving torque carries the same $sin θ$ , shrinking or growing the tilt scales both the push and the circle together — so the sweep rate cannot change.

Alt-text: side view, dotted vertical line, red L-vector tilted 30° from vertical, mint horizontal arrow labelled L sin theta showing the projection that is the radius of the precession circle, lavender arc marking the angle theta.

Verify: $0.784 rad/s$ — identical to Example 1's answer despite a totally different tilt. That is the $sin θ$ -cancellation the parent warned about, made concrete. ✓

C′ — over-tilt (θ beyond 90°)

Worked example Example 3 — Cell C′: the axis hangs

below horizontal Take the same wheel, but now the axis droops so it is tilted $θ = 12 0^{\circ}$ from vertical — i.e. the centre of mass end is $3 0^{\circ}$ below the horizontal. Compute $τ$ , $L sin θ$ , and $Ω$ . Does the precession rate change? Does its direction?

Forecast: past $9 0^{\circ}$ , does $sin θ$ go negative and flip everything?

$τ = r m g sin θ = 2.94 \times sin 12 0^{\circ} = 2.94 \times 0.866 = 2.546 N\cdotm$ . Why this step? $sin$ is positive for any angle between $0^{\circ}$ and $18 0^{\circ}$ , so a physical tilt (which never exceeds $18 0^{\circ}$ ) always gives a positive torque magnitude — nothing flips in the rate.
$L sin θ = 3.75 \times sin 12 0^{\circ} = 3.75 \times 0.866 = 3.248 kg\cdotm^{2} / s$ . Why this step? Same projection idea; $sin 12 0^{\circ} = sin 6 0^{\circ}$ , so the horizontal radius equals what it would be at $6 0^{\circ}$ .
$Ω = \frac{τ}{L sin θ} = \frac{2.546}{3.248} = 0.784 rad/s$ . Why this step? The $sin θ$ cancels again — the rate is still $0.784$ , tilt-independent all the way from just above $0^{\circ}$ to just below $18 0^{\circ}$ .
What DOES change is the sense of the swing: with the mass end now below horizontal, the horizontal component of $r$ has reversed, so $τ = r \times m g$ points the opposite way, and the axis precesses in the opposite rotational sense compared to the $6 0^{\circ}$ mirror case. Why this step? $sin θ$ controls the magnitude; the direction is set by where $r$ actually points, which the cross product tracks honestly (Cross Product and Right-Hand Rule).

Verify: $Ω = 0.784 rad/s$ , identical to both earlier cases — confirming tilt-independence holds across the whole $0 < θ < 18 0^{\circ}$ range, while the direction is the only thing over-tilt reverses. ✓

D — the fast-spin limit

E — the vertical degenerate case

Worked example Example 5 — Cell E: axis dead vertical,

$θ = 0$ The same wheel, but its spin axis points straight up ( $θ = 0^{\circ}$ ). What is the torque and what happens?

Forecast: with the axle vertical, where does gravity's lever arm go?

$τ = r m g sin θ = 2.94 \times sin 0^{\circ} = 2.94 \times 0 = 0 N\cdotm$ . Why this step? $r$ (up the axis) and $g$ (down) are anti-parallel; the cross product of parallel/anti-parallel vectors is zero.
With $τ = 0$ , $d L = τ d t = 0$ . The axis does not move. Why this step? No torque ⇒ no change in $L$ ⇒ no precession.
Reading the $0/0$ properly. The compact form $Ω = τ / (L sin θ)$ has $τ = r m g sin θ$ , so it is really $Ω = \frac{r m g sin θ}{L sin θ}$ . Taking the limit as the tilt shrinks to zero, $θ \to 0 lim \frac{r m g sin θ}{L sin θ} = \frac{r m g}{L}$ — a finite number, not zero and not infinity. So the algebra never actually blows up; the honest limit is the same $r m g / L = 0.784 rad/s$ . Why this step? The $0/0$ is only apparent — cancel the shared $sin θ$ before substituting. The genuine physical statement is separate: at exactly $θ = 0$ the torque is literally zero, so although the rate formula has a well-defined limit, there is nothing to precess because $d L = 0$ . Both facts are true at once.

Verify: Physically a perfectly upright spinning top ("sleeping top") sits still — matches $τ = 0$ . The limiting rate $r m g / L = 0.784$ is finite, so the singularity was removable; the top simply has no torque to start it precessing. ✓

F — the not-spinning degenerate case

Worked example Example 6 — Cell F: what if it isn't spinning?

Set $ω = 0$ (a dead, non-spinning wheel) held at $θ = 4 5^{\circ}$ . What does $Ω = r m g / (I ω)$ predict, and what really happens?

Forecast: plug $ω = 0$ into the denominator. Alarm bells?

$Ω = \frac{r m g}{I \cdot 0} = \frac{2.94}{0} \to \infty$ . Why this step? Dividing by zero flags a breakdown: the precession picture assumed a spin to steer.
Correct physics: with $L = 0$ there is no angular momentum to reorient. The torque instead produces angular acceleration — the wheel just falls/topples about the pivot, $α = τ / I$ . Why this step? $τ = d L / d t$ still holds; with $L$ starting at $0$ the torque builds $L$ in its own direction (tipping), rather than merely rotating an existing $L$ .
Numerically the initial angular acceleration is $α = \frac{τ}{I} = \frac{2.94}{0.125} = 23.52 rad/s^{2}$ . Why this step? This is ordinary rotational Newton's law — no precession, pure toppling.

Verify: $Ω \to \infty$ is the formula's way of saying "this regime doesn't apply." A non-spinning wheel obviously falls, giving finite $α = 23.52 rad/s^{2}$ . ✓

G — real-world word problem (mixed units)

Worked example Example 7 — Cell G: a gyrocompass rotor with messy units

A navigation gyroscope rotor is a solid disc ( $I = \frac{1}{2} M R^{2}$ ) with mass $M = 500 g$ and diameter $D = 8.0 cm$ , spinning at $N = 11, 459 rpm$ . A tiny drift torque $τ = 15 mN\cdotm$ (that's milli-newton-metres) acts on it. Find the drift precession rate $Ω$ and how long it takes to drift through one full circle. Convert every unit to SI first.

Forecast: high spin + tiny torque — expect a very slow drift.

Convert the units. Mass $M = 500 g = 0.50 kg$ . Radius $R = \frac{D}{2} = \frac{8.0 cm}{2} = 4.0 cm = 0.040 m$ . Spin $ω = N \times \frac{2 π}{60} = 11, 459 \times \frac{2 π}{60} \approx 1200 rad/s$ . Torque $τ = 15 mN\cdotm = 15 \times 1 0^{- 3} N\cdotm = 1.5 \times 1 0^{- 2} N\cdotm$ . Why this step? The formula only accepts SI (kg, m, rad/s, N·m); grams, cm, rpm and mN·m must be converted or the answer is nonsense. This is the whole point of a "mixed units" problem.
$I = \frac{1}{2} M R^{2} = \frac{1}{2} (0.50) (0.040)^{2} = 4.0 \times 1 0^{- 4} kg\cdotm^{2}$ . Why this step? A solid disc's moment of inertia is $\frac{1}{2} M R^{2}$ (Moment of Inertia).
$L = I ω = 4.0 \times 1 0^{- 4} \times 1200 = 0.48 kg\cdotm^{2} / s$ . Why this step? Spin angular momentum of the rotor.
$Ω = \frac{τ}{L} = \frac{1.5 \times 1 0 ^{- 2}}{0.48} = 3.125 \times 1 0^{- 2} rad/s$ . Why this step? Here $τ$ is given directly (a stray torque), so use $Ω = τ / L$ .
$T = \frac{2 π}{Ω} = \frac{2 π}{3.125 \times 1 0 ^{- 2}} \approx 201 s \approx 3.35 min$ . Why this step? Full-circle drift time is $2 π$ divided by the sweep rate.

Verify: all quantities landed in SI, and a small torque against a decent $L$ gives a slow drift of a few hundredths of a rad/s, one lap in ~3.4 min — the kind of slow, predictable drift a real gyrocompass must correct for (Gyroscope and Navigation). Units of $Ω$ : $\frac{N\cdotm}{kg\cdotm ^{2} / s} = rad/s$ . ✓

H — exam twist (no calculator)

Worked example Example 8 — Cell H: scaling puzzle

A top precesses with period $T_{0} = 6 s$ . Without changing anything else you (a) triple the spin rate $ω$ , then separately (b) double the mass $m$ . Give the new period in each case, in your head.

Forecast: $Ω = r m g / (I ω)$ . Which knob helps, which hurts?

Recall $Ω \propto \frac{m}{ω}$ and $T = \frac{2 π}{Ω} \propto \frac{ω}{m}$ . Why this step? Strip the formula to just the quantities being changed; treat $r, g, I$ as fixed constants. (For a fixed geometry, doubling $m$ also doubles $I$ — but the extra torque and extra inertia trade off in a way the problem asks us to treat with $I$ held fixed, i.e. mass added at the pivot end so only $τ$ grows.)
(a) Triple $ω$ ⇒ $Ω$ becomes $\frac{1}{3}$ , so $T$ triples: $T = 3 \times 6 = 18 s$ . Why this step? $ω$ in the denominator of $Ω$ ; period is inverse of $Ω$ .
(b) Double $m$ (with $I$ fixed) ⇒ $Ω$ doubles, so $T$ halves: $T = 6/2 = 3 s$ . Why this step? $m$ is in the numerator of $Ω$ (through $τ = r m g$ ); more torque ⇒ faster precession ⇒ shorter period.

Verify: faster spin → longer period (more stable), more mass → shorter period (more torque). The two knobs push opposite ways, consistent with $T \propto ω / m$ . ✓

Active recall

Recall Which cell breaks the formula, and how do you handle it?

$θ = 0$ (Cell E) → $τ = 0$ ; the $0/0$ is removable ( $lim_{θ \to 0} r m g sin θ / (L sin θ) = r m g / L$ is finite), but physically there is no torque to start precession.
$ω = 0$ (Cell F) → denominator zero, formula invalid; the object simply falls with $α = τ / I$ .

Recall Why do Examples 1, 2 and 3 give the

same $Ω$ ? Because $Ω = \frac{r m g sin θ}{L sin θ}$ and the $sin θ$ cancels for any $θ$ in $(0^{\circ}, 18 0^{\circ})$ — the rate is tilt-independent; only the direction of the swing changes past $9 0^{\circ}$ .

Recall Cell G unit-conversion checkpoint

Convert before substituting: $500 g \to 0.50 kg$ , diameter $8.0 cm \to R = 0.040 m$ , $11, 459 rpm \to\approx 1200 rad/s$ , $15 mN\cdotm \to 1.5 \times 1 0^{- 2} N\cdotm$ ; then $Ω \approx 0.03125 rad/s$ .