Gyroscope — steady precession derivation: Worked examples

2.1.24 · D3 · Physics › Analytical Mechanics › Gyroscope — steady precession derivation

Yeh page ek drill hall hai. Koi bhi formula aane se pehle, main har ek letter ka naam bataunga taaki jo reader sirf IS page par aaya ho woh pehli line se follow kar sake.

Definition Symbols ki cast (har ek, shabdon mein — pehle use se pehle define kiya gaya)

$m$ = top ki mass, kilograms mein. Yeh kitna bhaari hai.
$g$ = gravitational acceleration $= 9.8 m/s^{2}$ . Earth kitni zyada kheenchti hai.
$ℓ$ = pivot point $O$ se center of mass tak ki doori, axis ke saath, metres mein. "Lever length."
$θ$ = axis ka tilt angle seedha-upar se (vertical se). $θ = 0$ matlab bilkul seedha; $θ = 9 0^{\circ}$ matlab bilkul leta hua. Dekho Euler Angles.
$I_{3}$ = axial moment of inertia: top ko apni KHUD ki axis ke baare mein spin karna kitna mushkil hai. Dekho Moment of Inertia Tensor.
$I_{1}$ = transverse moment of inertia: poore top ko ek taraf jhukana kitna mushkil hai.
$ω_{s}$ = symmetry axis ke baare mein spin rate, radians per second mein. Yeh kitni tezi se ghoomta hai.
$Ω = \dot{ϕ}$ = precession rate: axis kitni tezi se vertical circle ke around chalta hai. Yahi hum solve karte hain.

Ab jab har symbol ka ek saada-sa matlab aur ek picture hai (jhuki hui axis, lever, walk-around), yeh do tools hain jo parent note ne banaye:

Dono formulas ke peeche ki picture — jhuki hui axis, gravity center of mass par kheeench rahi hai, resulting torque, aur axis ka sideways "walk" — yahan hai. Isko baar baar refer karte rehna; neeche ke har example mein yahi ek diagram hai.

Figure — Gyroscope — steady precession derivation

Ise left to right padho: blue arrow spin angular momentum $L_{s} = I_{3} ω_{s}$ hai leaning axis ke saath; yellow dot center of mass hai $ℓ$ doori par; red arrow gravity $m g$ hai seedha neeche kheeenchti hui; green circle tip par precession walk hai rate $Ω$ par; white arc tilt angle $θ$ hai.

Scenario matrix

Is topic ke har problem ka answer in cells mein se kisi ek mein milta hai. Neeche ke nau worked examples har ek cell(s) ke saath tagged hain jo woh cover karte hain, isliye milke poori table cover ho jaati hai.

Cell	Case class	Kya cheez tricky banati hai
A	Standard fast top	seedha $Ω = m g ℓ / I_{3} ω_{s}$ mein plug karo
B	$Ω$ ka sign / direction	kaun si taraf chalta hai — clockwise ya counter?
C	Limiting input: $ω_{s} \to \infty$	precession $\to 0$ (super-ziddi)
D	Limiting input: $ω_{s} \to$ small	discriminant negative ho jaata hai — koi steady precession nahi
E	Degenerate tilt $θ = 9 0^{\circ}$	axis horizontal; kya formula survive karta hai?
F	Degenerate tilt $θ = 0^{\circ}$ (sleeping top)	axis vertical; $sin θ = 0$ har jagah
G	Exact vs approximate — dono roots	fast & slow precession compare kiye
H	Real-world word problem	English → symbols mein translate karo, units rakho
I	Exam twist: hidden variable ke liye solve karo	$Ω$ diya hai, $ω_{s}$ ya $I_{3}$ nikalo

Worked examples

Example 1 — Cell A (standard fast top)

Worked example Dum-roti wala case

Ek gyroscope ka $I_{3} = 0.10 kg\cdotm^{2}$ hai, $ω_{s} = 40 rad/s$ par spin karta hai, pivot se $ℓ = 0.20$ m hang karta hai, mass $m = 2.0$ kg. Precession rate $Ω$ nikalo.

Forecast: Andaza lagao — kya $Ω$ bada number hoga (fast walk) ya chhota (slow walk)? Fast spinner ziddi hota hai, isliye $Ω$ chhota expect karo.

Step 1. Identify karo kaunsa formula use karna hai. Spin bada hai, isliye fast-top use karo. Yeh step kyun? Fast-top valid hai jab $(I_{3} ω_{s})^{2} ≫ 4 I_{1} m g ℓ cos θ$ . Bade $ω_{s}$ ke saath approximation safe aur bahut simple hai.

Step 2. Numerator compute karo $m g ℓ = 2.0 \cdot 9.8 \cdot 0.20 = 3.92$ . Yeh step kyun? $m g ℓ$ hi gravity torque ki magnitude hai $θ = 9 0^{\circ}$ par — "toppling effort" (figure mein red arrow ki turning power).

Step 3. Denominator compute karo $I_{3} ω_{s} = 0.10 \cdot 40 = 4.0$ . Yeh step kyun? $I_{3} ω_{s}$ spin angular momentum $L_{s}$ hai — blue arrow ki length, "stubbornness."

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

Verify: Units: $\frac{kg \cdot ( m/s ^{2} ) \cdot m}{kg\cdotm ^{2} \cdot ( 1/ s )} = \frac{kg\cdotm ^{2} / s ^{2}}{kg\cdotm ^{2} / s} = \frac{1}{s}$ ✓. Yeh rad/s hai. Chhota number jaise forecast tha; period $2 π /0.98 \approx 6.4$ s — ek aaram-se wala sweep. ✓

Example 2 — Cell B (direction/sign)

Worked example Yeh kaun si taraf chalta hai?

Example 1 wahi gyro. Spin angular momentum $L_{s}$ tilted axis ke saath baahir ki taraf point karta hai (spin se right-hand rule). Axis $+ x$ direction mein jhukti hai. Gravity torque hai $τ = r \times m g$ . Axis kaun si taraf (clockwise ya counter-clockwise, upar se dekha hua) precess karta hai, aur $Ω$ ka sign kya hai?

Forecast: Torque horizontal aur axis ke perpendicular hai. Andaza: tip $τ$ ki direction mein move karta hai.

Step 1. $r = ℓ \overset{e}{^}_{3}$ rakho $+ x$ ki taraf leaning, isliye iska horizontal part $+ x$ hai, vertical part $+ z$ . Gravity $m g = - m g \overset{z}{^}$ . Yeh step kyun? Cross product lene ke liye actual vectors chahiye; sirf picture sign par mislead kar sakti hai. Figure s02 in teen vectors ko explicitly draw karta hai.

Step 2. $τ = r \times m g$ . Horizontal $+ x$ piece crossed with $- \overset{z}{^}$ deta hai $\overset{x}{^} \times (- \overset{z}{^}) = + \overset{y}{^}$ . Yeh step kyun? Right-hand rule: $\overset{x}{^} \times \overset{z}{^} = - \overset{y}{^}$ , isliye $\overset{x}{^} \times (- \overset{z}{^}) = + \overset{y}{^}$ . Torque $+ y$ point karta hai — figure s02 mein green arrow.

Step 3. Kyunki $d L = τ d t$ , $L$ ka horizontal part (jo $+ x$ point karta hai) $+ y$ ki taraf nudge hota hai. Isliye axis $+ x$ se $+ y$ ki taraf rotate karta hai: upar se dekha hua counter-clockwise, $Ω > 0$ about $+ \overset{z}{^}$ . Yeh step kyun? Precession kuch nahi hai bas $L$ ki tip $τ$ ko chase karti hai. $Ω$ ki direction = $L_{h}$ se $τ$ ki taraf ki direction.

Verify: $τ = Ω \times L$ ke saath consistency check: $Ω \times L_{h}$ ko $+ \overset{y}{^}$ ke barabar hona chahiye. Lo $Ω = Ω \overset{z}{^}$ , $L_{h} = L_{h} \overset{x}{^}$ : $\overset{z}{^} \times \overset{x}{^} = + \overset{y}{^}$ ✓. Magnitude $Ω = m g ℓ / (I_{3} ω_{s}) = 0.98$ rad/s pehle jaisa. ✓

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

Worked example Infinitely ziddi top

Example 1 lo lekin spin badhao: $ω_{s} = 400$ , phir $4000 rad/s$ . $Ω$ ka kya hota hai?

Forecast: Zyada spin = zyada ziddi = ??? Change ki direction andaza lagao.

Step 1. $Ω = m g ℓ / (I_{3} ω_{s}) = 3.92/ (0.10 \cdot 400) = 0.098 rad/s$ . Yeh step kyun? Wahi fast-top formula; sirf $ω_{s}$ badla.

Step 2. $ω_{s} = 4000$ par: $Ω = 3.92/ (0.10 \cdot 4000) = 0.0098 rad/s$ . Yeh step kyun? $Ω \propto 1/ ω_{s}$ , isliye das-guna spin deta hai ek-daswa precession.

Step 3. Limit: jaise $ω_{s} \to \infty$ , $Ω \to 0$ . Yeh step kyun? Infinitely spun top essentially apni axis ko jagah par freeze kar deta hai — ideal "gyroscopic compass" jo hamesha ke liye ek fixed direction point karta hai.

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

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

Worked example Bahut slow spin: reality check

Thin top: $I_{1} = 0.012$ , $I_{3} = 0.006 kg\cdotm^{2}$ , $m = 0.5$ kg, $ℓ = 0.05$ m, $θ = 3 0^{\circ}$ . Wo minimum spin nikalo jiske neeche steady precession impossible hai.

Forecast: Ek threshold hona chahiye — usse neeche top steadily precess nahi kar sakta, use nutate ya girna padega. Andaza: tens of rad/s.

Step 1. Exact quadratic ke discriminant ko $\geq 0$ maango: $(I_{3} ω_{s})^{2} \geq 4 I_{1} m g ℓ cos θ$ . Yeh step kyun? $Ω_{\pm}$ real hain sirf tab jab square root ke andar ka content non-negative ho. Negative discriminant = koi real precession rate nahi = koi steady motion nahi. Dekho Nutation.

Step 2. Right side compute karo: $4 \cdot 0.012 \cdot 0.5 \cdot 9.8 \cdot 0.05 \cdot cos 3 0^{\circ} = 4 \cdot 0.012 \cdot 0.5 \cdot 9.8 \cdot 0.05 \cdot 0.8660 = 0.010184$ . Yeh step kyun? Hum saare known plug karte hain; $cos 3 0^{\circ} = 0.8660$ .

Step 3. Toh $I_{3} ω_{s} \geq 0.010184 = 0.10092$ , deta hai $ω_{s} \geq 0.10092/0.006 = 16.8 rad/s$ . Yeh step kyun? $ω_{s}$ isolate karne ke liye $I_{3}$ se divide karo.

Verify: Units: numerator $kg\cdotm^{2} \cdot kg \cdot (m/s^{2}) \cdot m = (kg\cdotm^{2} / s)^{2} = kg\cdotm^{2} / s$ ; $I_{3}$ (kg·m²) se divide karo → 1/s ✓. Numeric $16.8$ rad/s, tens mein jaise forecast tha. ✓

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

Worked example Axis bilkul flat

Example 1 ka bicycle wheel apni axle ke saath horizontal pakda gaya hai, $θ = 9 0^{\circ}$ . Fast-top se $Ω$ nikalo, aur check karo ki exact quadratic blow up nahi hoti.

Forecast: $θ = 9 0^{\circ}$ par torque maximum hai ( $m g ℓ sin 9 0^{\circ} = m g ℓ$ ). Kya $Ω$ bhi blow up hota hai? $sin θ$ cancellation yaad karo.

Step 1. Fast-top: $Ω = m g ℓ / (I_{3} ω_{s}) = 0.98$ rad/s — Example 1 se unchanged. Yeh step kyun? Torque se $sin θ$ aur horizontal $L$ se $sin θ$ parent derivation mein cancel ho gaye, isliye $Ω$ approximation mein angle-free hai.

Step 2. Exact quadratic: $cos θ$ coefficient $cos 9 0^{\circ} = 0$ ho jaata hai. Quadratic degenerate hokar linear equation ban jaati hai: $- I_{3} ω_{s} Ω + m g ℓ = 0$ . Yeh step kyun? Jab leading $Ω^{2}$ coefficient vanish ho jaata hai, "quadratic" linear mein collapse ho jaati hai — ek aisa case jise alag handle karna padta hai warna $/ cos θ$ zero se divide kar dega.

Step 3. Linear form solve karo: $Ω = m g ℓ / (I_{3} ω_{s}) = 0.98$ rad/s — exact aur fast-top answers $θ = 9 0^{\circ}$ par bilkul ek jaisi hain. Yeh step kyun? $cos θ = 0$ ke saath $I_{1}$ term bilkul drop ho jaata hai, isliye koi correction nahi; dono formulas literally agree karte hain.

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

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

Worked example Seedha "sleeping" top

Ek top bilkul seedha spin karta hai, $θ = 0^{\circ}$ . Theory kya kehti hai $Ω$ ke baare mein?

Forecast: Seedha matlab koi lean nahi, koi horizontal offset nahi, koi toppling torque nahi. Andaza: precession undefined ya zero hai.

Step 1. Torque magnitude $m g ℓ sin θ = m g ℓ sin 0^{\circ} = 0$ . Yeh step kyun? Center of mass pivot ke seedha upar hai, isliye gravity ka lever arm zero hai — precession drive karne ke liye koi torque nahi.

Step 2. Fast-top equation $m g ℓ sin θ = Ω I_{3} ω_{s} sin θ$ mein, har term $sin θ = 0$ carry karta hai. Equation padhti hai $0 = 0$ — $Ω$ indeterminate hai, kisi bhi value par force nahi kiya gaya. Yeh step kyun? $sin θ$ cancel nahi kar sakte jab yeh zero ho. "Steady precession" question vacuous ban jaata hai: koi bhi $Ω$ (including $0$ ) balance trivially satisfy karta hai.

Step 3. Physically top bas jagah par spin karta hai ("sleeping"). Stability ek alag sawaal hai jo Nutation decide karta hai: upright stable hai agar $(I_{3} ω_{s})^{2} > 4 I_{1} m g ℓ$ (the $θ = 0$ discriminant with $cos 0 = 1$ ). Yeh step kyun? Kya thoda-sa jhuka sleeping top wapis seedha hoga ya girega, yeh Cell D se generalised $θ = 0$ par discriminant chahiye.

Verify: Example-4 top ke liye $θ = 0$ par discriminant: $4 I_{1} m g ℓ \cdot 1 = 4 \cdot 0.012 \cdot 0.5 \cdot 9.8 \cdot 0.05 = 0.01176$ ; chahiye $(I_{3} ω_{s})^{2} > 0.01176$ , yaani $ω_{s} > 0.01176 /0.006 = 18.1$ rad/s. $θ = 3 0^{\circ}$ case (16.8) se thoda zyada threshold, kyunki $cos 0 > cos 3 0^{\circ}$ ✓.

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

Worked example Ek saath do precession rates

Top with $I_{1} = 0.012$ , $I_{3} = 0.006 kg\cdotm^{2}$ , $m = 0.5$ kg, $ℓ = 0.05$ m, $θ = 3 0^{\circ}$ , $ω_{s} = 50 rad/s$ par spun (16.8 threshold se kaafi upar). Dono exact precession rates $Ω_{+}$ aur $Ω_{-}$ nikalo.

Forecast: Ek root familiar slow precession hona chahiye; doosra ek fast wala. Andaza lagao kaunsa $\pm$ ka sign slow wala deta hai.

Step 1. $I_{3} ω_{s} = 0.006 \cdot 50 = 0.30$ compute karo. Yeh step kyun? Yeh shared numerator term aur root ke andar leading term hai.

Step 2. Discriminant $(I_{3} ω_{s})^{2} - 4 I_{1} m g ℓ cos θ = 0.3 0^{2} - 0.010184 = 0.09 - 0.010184 = 0.079816$ ; iska root hai $0.28252$ . Yeh step kyun? Hum wahi $4 I_{1} m g ℓ cos θ = 0.010184$ reuse karte hain Example 4 se (wahi top, wahi angle).

Step 3. Denominator $2 I_{1} cos θ = 2 \cdot 0.012 \cdot 0.8660 = 0.020785$ . Yeh step kyun? Yeh dono roots ko scale karta hai.

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). Yeh step kyun? $-$ root, numerator mein near-cancellation se, small (physically usual) precession deta hai; $+$ rarely-seen fast branch deta hai.

Verify: $Ω_{-}$ ke liye fast-top prediction: $m g ℓ / (I_{3} ω_{s}) = (0.5 \cdot 9.8 \cdot 0.05) /0.30 = 0.245/0.30 = 0.817$ rad/s — exact $0.841$ ke $3%$ andar ✓. Dono roots quadratic $I_{1} Ω^{2} cos θ - I_{3} ω_{s} Ω + m g ℓ$ mein plug karne par $\approx 0$ dete hain. Dekho Symmetric Top aur Lagrangian Mechanics. ✓

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

Worked example Ship ka gyrocompass

Ek gyrocompass rotor ek solid disk hai, mass $m = 6.0$ kg, radius $R = 0.15$ m, $ω_{s} = 300 rad/s$ par spun. Iska center of mass pivot se $ℓ = 0.010$ m offset hai ek chhote pendulous weight ki wajah se (yahi cheez ise north dhundhne par majboor karti hai). Iska precession rate nikalo.

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

Step 1. Solid disk ka apni axis ke baare mein axial moment: $I_{3} = \frac{1}{2} m R^{2} = \frac{1}{2} \cdot 6.0 \cdot 0.1 5^{2} = 0.0675 kg\cdotm^{2}$ . Yeh step kyun? Hum "solid disk" ko uske moment of inertia mein translate karte hain; axial value $\frac{1}{2} m R^{2}$ standard result use karta hai.

Step 2. Numerator $m g ℓ = 6.0 \cdot 9.8 \cdot 0.010 = 0.588$ . Yeh step kyun? Pendulous offset $ℓ$ lever arm hai jo aligning torque produce karta hai.

Step 3. $Ω = m g ℓ / (I_{3} ω_{s}) = 0.588/ (0.0675 \cdot 300) = 0.588/20.25 = 0.02904 rad/s$ . Yeh step kyun? Fast-top yahan overwhelmingly apply hota hai ( $ω_{s}$ bahut bada, $ℓ$ bahut chhota), isliye hum toppling torque ko spin angular momentum se divide karte hain.

Verify: Units 1/s ✓. Period $2 π /0.02904 = 216$ s ≈ 3.6 minutes per revolution — slow, jaise forecast tha, aur exactly yahi wajah hai ki gyrocompasses steady heading dete hain. ✓

Example 9 — Cell I (exam twist: hidden variable ke liye solve karo)

Worked example Spin ke liye back-solve karo

Ek exam observed precession deta hai: ek gyro period $T = 4.0$ s ke saath precess karta hai. Iska data: $I_{3} = 0.050 kg\cdotm^{2}$ , $m = 1.5$ kg, $ℓ = 0.12$ m. Woh spin $ω_{s}$ nikalo jo iske paas hona chahiye.

Forecast: $Ω = m g ℓ / (I_{3} ω_{s})$ ko rearrange karo $ω_{s}$ isolate karne ke liye. Kuch tens of rad/s expect karo.

Step 1. Period ko $Ω$ mein convert karo: $Ω = 2 π / T = 2 π /4.0 = 1.5708 rad/s$ . Yeh step kyun? Formula angular rate $Ω$ use karta hai, period nahi; $Ω = 2 π / T$ ek revolution ki definition se.

Step 2. Rearrange karo: $ω_{s} = m g ℓ / (I_{3} Ω)$ . Yeh step kyun? Algebra — dono sides ko $ω_{s}$ se multiply karo aur unknown ke liye solve karne ke liye $I_{3} Ω$ se divide karo.

Step 3. Numerator $m g ℓ = 1.5 \cdot 9.8 \cdot 0.12 = 1.764$ . Phir $ω_{s} = 1.764/ (0.050 \cdot 1.5708) = 1.764/0.078540 = 22.46 rad/s$ . Yeh step kyun? Rearrange hone ke baad seedha substitution.

Verify: Wapis plug karo: $Ω = m g ℓ / (I_{3} ω_{s}) = 1.764/ (0.050 \cdot 22.46) = 1.764/1.123 = 1.571$ rad/s → period $2 π /1.571 = 4.0$ s ✓. Round-trip close ho jaata hai. ✓

Recall Quick self-test (answers cover karo)

Fast top ke liye kaunsa formula? ::: $Ω = m g ℓ / (I_{3} ω_{s})$ Kya fast-top limit mein $Ω$ tilt $θ$ par depend karta hai? ::: Nahi — $sin θ$ cancel ho jaata hai. $ω_{s} \to \infty$ hone par $Ω$ ka kya hota hai? ::: $Ω \to 0$ (axis freeze ho jaati hai). Steady precession ke exist hone ki condition? ::: $(I_{3} ω_{s})^{2} \geq 4 I_{1} m g ℓ cos θ$ $θ = 9 0^{\circ}$ par exact aur fast-top kaise compare karte hain? ::: Woh bilkul agree karte hain ( $cos 9 0^{\circ} = 0$ ). $θ = 0^{\circ}$ par torque kya hai? ::: Zero — precession indeterminate hai (sleeping top).