Gyroscopic effect — precession of spinning top: Worked examples

1.5.16 · D3 · Physics › Rotational Mechanics › Gyroscopic effect — precession of spinning top

Yeh page parent topic ka drill hall hai. Hum ek hi master formula lete hain aur use har kone mein push karte hain: signs, degenerate inputs, limits, ek word problem, aur ek exam trap. Yahan jo bhi hai woh parent se contradict nahi karta — hum bas wider jaate hain.

Koi bhi number se pehle: neeche har symbol parent mein earn kiya gaya hai. Agar koi word ( $L$ , torque, cross product) unfamiliar lage, pehle linked note kholein — yeh page assume karta hai ki aap formula ko ek sentence ki tarah padh sakte hain.

Scenario matrix

Is topic ka har problem inheen cells mein se ek hoga. Neeche har worked example us cell ke saath tagged hai jis par woh land karta hai, taaki mil ke poora grid cover ho jaaye.

Cell	Kya vary karta hai	Kahan kaata hai
A. Standard positive	sab quantities ordinary & positive	baseline number-crunch
B. Direction / sign	precession kis direction mein hoga? ( $r \times g$ ka quadrant)	Cross Product and Right-Hand Rule
C. Tilt cancellation	$θ$ badlo, dekho $Ω$ fixed rahta hai	$sin θ$ trap
C′. Over-tilt	$θ > 9 0^{\circ}$ (axis horizontal ke neeche jhuk jaata hai)	swing ka sign, rate ka nahi
D. Limit: fast spin	$ω \to \infty$	$Ω \to 0$ (super-stable)
E. Degenerate: $θ = 0$	axis bilkul vertical	$τ = 0$ , koi precession nahi
F. Degenerate: $ω = 0$	spin nahi kar raha	formula blow up karta hai → woh bas gir jaata hai
G. Word problem	real gyroscope, mixed units	navigation-style setup
H. Exam twist	scaling / ratio, koi calculator nahi	proportional reasoning

Worked examples

A + B — baseline number AUR uski direction

Worked example Example 1 — Cell A & B: woh wheel jo circle mein chalta hai

Ek ring-jaisa wheel: $m = 2 kg$ , radius $R = 0.25 m$ , isliye $I = m R^{2}$ . Yeh $ω = 30 rad/s$ par spin karta hai aur ek cord se hang karta hai jiske saath centre of mass pivot se $r = 0.15 m$ door hai. Tilt $θ = 9 0^{\circ}$ (axle horizontal). $Ω$ aur precession ki direction dhundho agar $L$ $+ x$ along point kare aur gravity $- z$ ho.

Forecast: guess karo — kya $Ω$ $1 rad/s$ se bada hoga ya chota? Free end kis taraf swing karega?

$I = m R^{2} = 2 (0.25)^{2} = 0.125 kg\cdotm^{2}$ . Yeh step kyun? Ek ring apna saara mass radius $R$ par rakhti hai; ring ka Moment of Inertia $m R^{2}$ hota hai.
$L = I ω = 0.125 \times 30 = 3.75 kg\cdotm^{2} / s$ . Yeh step kyun? $L = I ω$ spin angular momentum ki definition hai (Angular Momentum).
$τ = r m g = 0.15 \times 2 \times 9.8 = 2.94 N\cdotm$ . Yeh step kyun? Gravity centre of mass par act karti hai; pivot ke baare mein torque (arm)×(force) hai, axle horizontal hone ki wajah se $sin θ = 1$ .
$Ω = \frac{τ}{L} = \frac{2.94}{3.75} = 0.784 rad/s$ . Yeh step kyun? Seedha master formula mein plug in karo.
Direction. $τ = r \times m g$ . $r$ ke saath $+ x$ along aur $g$ ke saath $- z$ along: $\overset{x}{^} \times (- \overset{z}{^}) = + \overset{y}{^}$ (kyunki $\overset{x}{^} \times \overset{z}{^} = - \overset{y}{^}$ ). Toh $d L ∥ + y$ : free tip $+ x$ se $+ y$ ki taraf swing karta hai — upar se dekha jaaye toh counter-clockwise. Yeh step kyun? $L$ ke change ki direction hamesha $τ$ ki direction hoti hai; kabhi guess mat karo, cross product compute karo (Cross Product and Right-Hand Rule, aur Vector nature of dL/dt bhi dekho).

Figure — Fig s01 (swing ka top view). Yeh picture woh kaam karta hai jo words nahi kar sakte: $L$ aur $τ$ ko literally $9 0^{\circ}$ par rakhta hai taaki aap dekh sako ki push sideways hai, downward nahi. Origin par pivot; red arrow $L$ $+ x$ along; mint arrow $τ$ (isliye $d L$ ) $+ y$ ki taraf; butter-yellow arc = tip counter-clockwise curl karke us dashed lavender circle par jaata hai jise woh trace karega. Pedagogy yeh hai: jo perpendicularity precession ko counter-intuitive banati hai woh spatial hai, isliye sirf ek spatial figure hi usse settle karta hai.

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: $Ω < 1$ , jo forecast se match karta hai ki ek decent spin lazy precession deta hai. Units: $\frac{N\cdotm}{kg\cdotm ^{2} / s} = \frac{kg\cdotm ^{2} / s ^{2}}{kg\cdotm ^{2} / s} = s^{- 1} = rad/s$ . ✓ Aur $\overset{x}{^} \times (- \overset{z}{^}) = + \overset{y}{^}$ swing ko confirm karta hai.

C — tilt cancellation trap

Worked example Example 2 — Cell C: tilt badlo, rate rakho waisa hi

Example 1 waala same wheel ( $r m g = 2.94 N\cdotm$ , $L = 3.75$ ). Ab isse $θ = 3 0^{\circ}$ par tilt karo. $τ$ , horizontal- $L$ radius, aur $Ω$ compute karo. $9 0^{\circ}$ case se compare karo.

Forecast: torque drop karega (horizontal mein lever arm kam hai). Kya $Ω$ bhi drop karega?

$τ = r m g sin θ = 2.94 \times sin 3 0^{\circ} = 2.94 \times 0.5 = 1.47 N\cdotm$ . Yeh step kyun? Sirf gravity ka woh component jo axis ke perpendicular hai torque banata hai; woh $sin θ$ le aata hai.
$L$ ka horizontal projection magnitude $L sin θ = 3.75 \times 0.5 = 1.875 kg\cdotm^{2} / s$ hai. Yeh step kyun? Precession matlab $L$ ki tip vertical ke around jaana. $L$ ka sirf woh hissa jo horizontal plane mein hota hai actually circle karta hai; vertical part bas wahan baitha rehta hai. Toh horizontal projection literally woh circle ka radius hai jis par tip travel karti hai — figure dekho, jahan mint arrow $L sin θ$ tilted red $L$ ki flat shadow hai.
$Ω = \frac{τ}{L sin θ} = \frac{1.47}{1.875} = 0.784 rad/s$ . Yeh step kyun? Do $sin θ$ factors — ek upar $τ$ mein, ek neeche projection mein — cancel ho jaate hain.

Figure — Fig s02 (kyun tilt cancel hota hai). Figure apni jagah earn karta hai kyunki yeh ek hi frame mein do cheezein dikhata hai jo dono $sin θ$ ke saath scale karti hain, toh cancellation algebraic ke bajaye visible ho jaati hai. Dotted line = vertical; red arrow = $L$ lavender angle $θ = 3 0^{\circ}$ se tilt hua; flat mint arrow horizontal par gira hua, length $L sin θ$ , yeh precession circle ka radius hai. Kyunki driving torque mein bhi wahi $sin θ$ hai, tilt ko ghata-badha kar dono push aur circle ek saath scale hote hain — isliye sweep rate change nahi ho sakti.

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: $0.784 rad/s$ — Example 1 ke answer se bilkul same, bilkul alag tilt ke bawajood. Yahi woh $sin θ$ -cancellation hai jiske baare mein parent ne warn kiya tha, ab concrete ho gayi. ✓

C′ — over-tilt (θ 90° se aage)

Worked example Example 3 — Cell C′: axis horizontal ke

neeche jhuk jaata hai Wohi wheel lo, lekin ab axis itna jhuk jaata hai ki vertical se $θ = 12 0^{\circ}$ tilt ho jaata hai — yaani centre of mass wala end horizontal se $3 0^{\circ}$ neeche hai. $τ$ , $L sin θ$ , aur $Ω$ compute karo. Kya precession rate change hoti hai? Kya direction?

Forecast: $9 0^{\circ}$ ke baad, kya $sin θ$ negative ho jaata hai aur sab kuch flip karta hai?

$τ = r m g sin θ = 2.94 \times sin 12 0^{\circ} = 2.94 \times 0.866 = 2.546 N\cdotm$ . Yeh step kyun? $sin$ $0^{\circ}$ aur $18 0^{\circ}$ ke beech positive rehta hai, isliye ek physical tilt (jo kabhi $18 0^{\circ}$ se zyada nahi hoti) hamesha positive torque magnitude deta hai — rate mein kuch flip nahi hota.
$L sin θ = 3.75 \times sin 12 0^{\circ} = 3.75 \times 0.866 = 3.248 kg\cdotm^{2} / s$ . Yeh step kyun? Wohi projection idea; $sin 12 0^{\circ} = sin 6 0^{\circ}$ , isliye horizontal radius $6 0^{\circ}$ par hone jaisi hogi.
$Ω = \frac{τ}{L sin θ} = \frac{2.546}{3.248} = 0.784 rad/s$ . Yeh step kyun? $sin θ$ phir cancel ho jaata hai — rate ab bhi $0.784$ hai, $0^{\circ}$ se bilkul theek $18 0^{\circ}$ tak tilt-independent.
Jo BADALTA HAI woh swing ka sense hai: mass end ab horizontal ke neeche hone se, $r$ ka horizontal component reverse ho gaya hai, isliye $τ = r \times m g$ opposite direction mein point karta hai, aur axis $6 0^{\circ}$ mirror case ke compare mein opposite rotational sense mein precess karta hai. Yeh step kyun? $sin θ$ magnitude control karta hai; direction is baat se set hoti hai ki $r$ actually kahan point kar raha hai, jise cross product honestly track karta hai (Cross Product and Right-Hand Rule).

Verify: $Ω = 0.784 rad/s$ , teeno earlier cases se same — confirm karta hai ki tilt-independence poore $0 < θ < 18 0^{\circ}$ range mein hold karta hai, jabki direction woh akela cheez hai jise over-tilt reverse karta hai. ✓

D — fast-spin limit

E — vertical degenerate case

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

$θ = 0$ Wohi wheel, lekin uska spin axis seedha upar point karta hai ( $θ = 0^{\circ}$ ). Torque kya hai aur kya hota hai?

Forecast: axle vertical hone par, gravity ka lever arm kahan jaata hai?

$τ = r m g sin θ = 2.94 \times sin 0^{\circ} = 2.94 \times 0 = 0 N\cdotm$ . Yeh step kyun? $r$ (axis ke upar) aur $g$ (neeche) anti-parallel hain; parallel/anti-parallel vectors ka cross product zero hota hai.
$τ = 0$ ke saath, $d L = τ d t = 0$ . Axis nahi hilta. Yeh step kyun? Koi torque nahi ⇒ $L$ mein koi change nahi ⇒ koi precession nahi.
$0/0$ ko theek se padhna. Compact form $Ω = τ / (L sin θ)$ mein $τ = r m g sin θ$ hai, isliye yeh actually $Ω = \frac{r m g sin θ}{L sin θ}$ hai. Jaise tilt zero tak shrink hota hai, $θ \to 0 lim \frac{r m g sin θ}{L sin θ} = \frac{r m g}{L}$ — yeh ek finite number hai, na zero na infinity. Toh algebra actually kabhi blow up nahi karta; honest limit wohi $r m g / L = 0.784 rad/s$ hai. Yeh step kyun? $0/0$ sirf apparent hai — substitute karne se pehle shared $sin θ$ cancel karo. Genuine physical statement alag hai: exactly $θ = 0$ par torque literally zero hai, isliye although rate formula ka well-defined limit hai, precession ke liye kuch bhi nahi kyunki $d L = 0$ . Dono facts ek saath sach hain.

Verify: Physically ek bilkul seedha spinning top ("sleeping top") still baitha rehta hai — $τ = 0$ se match karta hai. Limiting rate $r m g / L = 0.784$ finite hai, isliye singularity removable thi; top ke paas bas koi torque nahi hai precession shuru karne ke liye. ✓

F — not-spinning degenerate case

Worked example Example 6 — Cell F: agar yeh spin nahi kar raha toh?

$ω = 0$ set karo (ek dead, non-spinning wheel) $θ = 4 5^{\circ}$ par hold kiya gaya. $Ω = r m g / (I ω)$ kya predict karta hai, aur actually kya hota hai?

Forecast: $ω = 0$ denominator mein plug karo. Alarm bells?

$Ω = \frac{r m g}{I \cdot 0} = \frac{2.94}{0} \to \infty$ . Yeh step kyun? Zero se divide karna ek breakdown flag karta hai: precession picture ne assume kiya tha ki steer karne ke liye spin hogi.
Correct physics: $L = 0$ ke saath koi angular momentum nahi hai reorient karne ke liye. Torque instead angular acceleration produce karta hai — wheel bas gir/topple karta hai pivot ke baare mein, $α = τ / I$ . Yeh step kyun? $τ = d L / d t$ abhi bhi hold karta hai; $L$ ke $0$ se start karne ke saath torque apni direction mein $L$ build karta hai (tipping), naa ki existing $L$ ko merely rotate karta hai.
Numerically initial angular acceleration hai $α = \frac{τ}{I} = \frac{2.94}{0.125} = 23.52 rad/s^{2}$ . Yeh step kyun? Yeh ordinary rotational Newton's law hai — koi precession nahi, pure toppling.

Verify: $Ω \to \infty$ formula ka tarika hai yeh kehne ka ki "yeh regime apply nahi hota." Ek non-spinning wheel obviously girta hai, finite $α = 23.52 rad/s^{2}$ deta hai. ✓

G — real-world word problem (mixed units)

Worked example Example 7 — Cell G: gyrocompass rotor messy units ke saath

Ek navigation gyroscope rotor solid disc hai ( $I = \frac{1}{2} M R^{2}$ ) mass $M = 500 g$ aur diameter $D = 8.0 cm$ ke saath, $N = 11, 459 rpm$ par spin karta hai. Ek tiny drift torque $τ = 15 mN\cdotm$ (yeh milli-newton-metres hai) uss par act karta hai. Drift precession rate $Ω$ nikalo aur kitne time mein ek poora circle drift karta hai. Pehle har unit ko SI mein convert karo.

Forecast: high spin + tiny torque — expect karo ek bahut slow drift.

Units convert karo. 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 $τ = 15 mN\cdotm = 15 \times 1 0^{- 3} N\cdotm = 1.5 \times 1 0^{- 2} N\cdotm$ . Yeh step kyun? Formula sirf SI accept karta hai (kg, m, rad/s, N·m); grams, cm, rpm aur mN·m convert karne honge warna answer bakwaas hoga. Yahi "mixed units" problem ka poora point hai.
$I = \frac{1}{2} M R^{2} = \frac{1}{2} (0.50) (0.040)^{2} = 4.0 \times 1 0^{- 4} kg\cdotm^{2}$ . Yeh step kyun? Solid disc ka moment of inertia $\frac{1}{2} M R^{2}$ hota hai (Moment of Inertia).
$L = I ω = 4.0 \times 1 0^{- 4} \times 1200 = 0.48 kg\cdotm^{2} / s$ . Yeh step kyun? Rotor ka spin angular momentum.
$Ω = \frac{τ}{L} = \frac{1.5 \times 1 0 ^{- 2}}{0.48} = 3.125 \times 1 0^{- 2} rad/s$ . Yeh step kyun? Yahan $τ$ directly diya gaya hai (ek stray torque), isliye $Ω = τ / L$ use karo.
$T = \frac{2 π}{Ω} = \frac{2 π}{3.125 \times 1 0 ^{- 2}} \approx 201 s \approx 3.35 min$ . Yeh step kyun? Full-circle drift time $2 π$ ko sweep rate se divide karna hai.

Verify: saari quantities SI mein aayi, aur ek decent $L$ ke against ek chota torque slow drift deta hai kuch hundredths of rad/s, ek lap ~3.4 min mein — isi tarah ka slow, predictable drift hai jise ek real gyrocompass ko correct karna padta hai (Gyroscope and Navigation). $Ω$ ke units: $\frac{N\cdotm}{kg\cdotm ^{2} / s} = rad/s$ . ✓

H — exam twist (no calculator)

Worked example Example 8 — Cell H: scaling puzzle

Ek top period $T_{0} = 6 s$ ke saath precess karta hai. Baaki kuch change kiye bina aap (a) spin rate $ω$ ko triple karte ho, phir alag se (b) mass $m$ ko double karte ho. Dono cases mein naya period apne dimaag mein do.

Forecast: $Ω = r m g / (I ω)$ . Kaunsa knob help karta hai, kaunsa hurt?

Yaad karo $Ω \propto \frac{m}{ω}$ aur $T = \frac{2 π}{Ω} \propto \frac{ω}{m}$ . Yeh step kyun? Formula ko sirf un quantities tak strip karo jo change ho rahi hain; $r, g, I$ ko fixed constants maano. (Fixed geometry ke liye, $m$ double karne se $I$ bhi double hota hai — lekin extra torque aur extra inertia us tarike se trade off karte hain jaise problem humse $I$ ko fixed maan ke treat karne ko kehti hai, yaani mass pivot end par add kiya gaya taaki sirf $τ$ badhey.)
(a) $ω$ triple karo ⇒ $Ω$ $\frac{1}{3}$ ho jaata hai, isliye $T$ triple hoti hai: $T = 3 \times 6 = 18 s$ . Yeh step kyun? $ω$ $Ω$ ke denominator mein hai; period $Ω$ ka inverse hai.
(b) $m$ double karo ( $I$ fixed ke saath) ⇒ $Ω$ double hoti hai, isliye $T$ half ho jaati hai: $T = 6/2 = 3 s$ . Yeh step kyun? $m$ $Ω$ ke numerator mein hai ( $τ = r m g$ ke through); zyada torque ⇒ faster precession ⇒ shorter period.

Verify: faster spin → longer period (more stable), zyada mass → shorter period (zyada torque). Do knobs opposite directions mein push karte hain, $T \propto ω / m$ se consistent. ✓

Active recall

Recall Kaunsa cell formula ko break karta hai, aur aap usse kaise handle karte ho?

$θ = 0$ (Cell E) → $τ = 0$ ; $0/0$ removable hai ( $lim_{θ \to 0} r m g sin θ / (L sin θ) = r m g / L$ finite hai), lekin physically precession shuru karne ke liye koi torque nahi hai.
$ω = 0$ (Cell F) → denominator zero, formula invalid; object bas $α = τ / I$ ke saath gir jaata hai.

Recall Examples 1, 2 aur 3 mein

same $Ω$ kyun aata hai? Kyunki $Ω = \frac{r m g sin θ}{L sin θ}$ aur $sin θ$ $(0^{\circ}, 18 0^{\circ})$ mein kisi bhi $θ$ ke liye cancel ho jaata hai — rate tilt-independent hai; $9 0^{\circ}$ ke baad sirf swing ki direction change hoti hai.

Recall Cell G unit-conversion checkpoint

Substitute karne se pehle convert karo: $500 g \to 0.50 kg$ , diameter $8.0 cm \to R = 0.040 m$ , $11, 459 rpm \to\approx 1200 rad/s$ , $15 mN\cdotm \to 1.5 \times 1 0^{- 2} N\cdotm$ ; phir $Ω \approx 0.03125 rad/s$ .