Gyroscope — steady precession derivation: Exercises

2.1.24 · D4 · Physics › Analytical Mechanics › Gyroscope — steady precession derivation

Neeche sab kuch ke liye quick reference (sab parent mein build hai):

Symbol reminder (plain words mein):

$m$ = top ki mass, $g = 9.8 m/s^{2}$ = gravitational field strength.
$ℓ$ = pivot $O$ se centre of mass tak ki distance, axis ke saath measure ki gayi.
$θ$ = axis ka seedha-upar (vertical) se jhukav. Euler Angles dekho.
$ω_{s}$ = top ki apni symmetry axis ke baare mein spin rate.
$I_{3}$ = symmetry axis ke baare mein moment of inertia; $I_{1}$ = $O$ se guzarne wali transverse axis ke baare mein. Symmetric Top dekho.
$Ω = \dot{ϕ}$ = precession rate (axis kitni tezi se vertical ke around ghoomti hai).

Intuition Shuru karne se pehle figure padho

Neeche ki picture woh geometry hai jo is page ke har problem mein use hoti hai. Ek baar trace karo: kali dotted vertical "seedha upar" ( $z$ ) ki direction hai. Teal arrow top ki axis hai, jisme spin angular momentum $L_{s}$ carry hoti hai, vertical se $θ$ angle (kala arc) se jhuki hui. Plum dot centre of mass hai jo axis ke saath $ℓ$ distance par hai; pivot se guzarne wali vertical line se uska horizontal offset lever arm $ℓ sin θ$ hai (kala double arrow). Orange arrow gravity $m g$ hai jo CM par seedha neeche khich rahi hai. Unka cross product $τ = r \times m g$ circle-with-a-dot hai (ek vector page se bahar point karta hua): yeh horizontal aur axis ke perpendicular hai, aur isi liye yeh $L$ ko side ki taraf sweep karta hai bajaye girne ke. Har solution ke liye yeh image dimag mein rakho.

Figure — Gyroscope — steady precession derivation

Intuition

$Ω$ ke liye sign convention (poori jagah use hota hai) Hum $ϕ$ (aur isliye $Ω = \dot{ϕ}$ ) ko upward vertical $+ \overset{z}{^}$ ke around rotation ke roop mein measure karte hain, right-hand rule se positive (upar se dekha jaaye to counter-clockwise). Is convention se:

$cos θ > 0$ (top horizontal se upar jhuka, $θ < 9 0^{\circ}$ ): fast-top formula $Ω > 0$ deta hai — ordinary (prograde) precession, axis upar se dekha jaaye to counter-clockwise chalti hai.
$cos θ < 0$ ( $θ > 9 0^{\circ}$ , CM pivot ke neeche latka hai): denominator $2 I_{1} cos θ$ ka sign flip ho jaata hai, isliye roots ka sign badal jaata hai — axis ulti taraf sweep hoti hai (retrograde precession). Hum yeh L5 mein explicitly treat karte hain. Ek negative root koi galti nahi hai: iska matlab bas yeh hai "upar se dekha jaaye to clockwise precesses." Uska magnitude phir bhi ek physical rate hai.

Level 1 — Recognition

L1.1 Ek gyroscope mein $I_{3} = 0.10 kg\cdotm^{2}$ , spin $ω_{s} = 40 rad/s$ , mass $m = 2.0$ kg, aur pivot-to-CM distance $ℓ = 0.20$ m hai. Fast-top precession rate $Ω$ nikalo.

Recall Solution

KYA: seedha fast-top formula mein plug karo. KYUN: wheel tezi se ghoom raha hai, isliye simple formula apply hota hai. $Ω = \frac{m g ℓ}{I _{3} ω _{s}} = \frac{2.0 \cdot 9.8 \cdot 0.20}{0.10 \cdot 40} = \frac{3.92}{4.0} = 0.98 rad/s .$ Axis $0.98$ rad/s par ek slow horizontal circle sweep karti hai (prograde, kyunki $θ < 9 0^{\circ}$ ).

L1.2 Usi gyroscope ke liye, ek poori precession sweep kitne time mein hoti hai?

Recall Solution

KYA: angular rate ko period mein convert karo. KYUN: ek poora circle $2 π$ radians hota hai, isliye time $= 2 π /Ω$ . $T = \frac{2 π}{Ω} = \frac{2 π}{0.98} \approx 6.41 s .$

L1.3 Sahi ya galat: fast-top approximation mein, axis ko steeper angle $θ$ par jhukane se woh tezi se precess karta hai. Ek sentence mein justify karo.

Recall Solution

Galat. Fast-top $Ω = m g ℓ / (I_{3} ω_{s})$ mein $θ$ bilkul nahi aata — torque se $sin θ$ aur rotating horizontal Angular Momentum se $sin θ$ exactly cancel ho jaate hain.

Level 2 — Application

L2.1 Friction L1 wheel ki spin ko $ω_{s} = 20 rad/s$ tak slow kar deta hai. Naya $Ω$ kya hai?

Recall Solution

KYA: formula ko chhoti spin ke saath re-apply karo. KYUN: $Ω \propto 1/ ω_{s}$ , isliye spin ko aadha karne se $Ω$ double ho jaata hai. $Ω = \frac{3.92}{0.10 \cdot 20} = \frac{3.92}{2.0} = 1.96 rad/s .$ Jaise real tops spin down hote hain, woh tezi se precess karte hain — theek wahi jo tum ek dying top ke wobble karne aur girne se pehle dekhte ho.

L2.2 Ek toy gyroscope $Ω = 2.5$ rad/s par precess karta hai. Uske parameters hain $m = 0.30$ kg, $ℓ = 0.040$ m, $I_{3} = 1.2 \times 1 0^{- 4} kg\cdotm^{2}$ . Uski spin rate $ω_{s}$ nikalo.

Recall Solution

KYA: fast-top formula ko $ω_{s}$ ke liye invert karo. KYUN: $Ω$ measure kiya gaya hai, baaki sab known hai, isliye algebraically solve karo. $ω_{s} = \frac{m g ℓ}{I _{3} Ω} = \frac{0.30 \cdot 9.8 \cdot 0.040}{1.2 \times 1 0 ^{- 4} \cdot 2.5} = \frac{0.11760}{3.0 \times 1 0 ^{- 4}} = 392 rad/s .$

L2.3 Do identical gyroscopes ek hi rate par spin karte hain, lekin gyroscope B ka centre of mass pivot se double dur hai ( $ℓ_{B} = 2 ℓ_{A}$ ). Kaun tezi se precess karta hai, aur kitne factor se?

Recall Solution

KYA: $Ω \propto ℓ$ (baaki sab equal) compare karo. KYUN: bada lever arm $ℓ$ zyada gravitational torque matlab hai, aur torque precession drive karta hai. $\frac{Ω _{B}}{Ω _{A}} = \frac{ℓ _{B}}{ℓ _{A}} = 2.$ Gyroscope B double tezi se precess karta hai.

Level 3 — Analysis

L3.1 Ek thin symmetric top mein $I_{1} = 0.012 kg\cdotm^{2}$ , $I_{3} = 0.006 kg\cdotm^{2}$ , $m = 0.5$ kg, $ℓ = 0.05$ m, tilt $θ = 3 0^{\circ}$ hai. Minimum spin $ω_{s}$ nikalo jiske liye steady precession exist kar sake.

Recall Solution

KYA: exact quadratic mein discriminant ko non-negative demand karo. KYUN: agar square root ka argument negative ho jaaye, $Ω$ complex ho jaata hai — koi real steady precession nahi, isliye top ko nutate karna padega ya girna padega. Require $(I_{3} ω_{s})^{2} \geq 4 I_{1} m g ℓ cos θ$ : $ω_{s} \geq \frac{4 I _{1} m g ℓ c o s θ}{I _{3}} = \frac{4 \cdot 0.012 \cdot 0.5 \cdot 9.8 \cdot 0.05 \cdot c o s 3 0 ^{\circ}}{0.006} .$ Root ke andar: $4 \cdot 0.012 \cdot 0.5 \cdot 9.8 \cdot 0.05 \cdot 0.8660 = 0.010184$ . Isliye $ω_{s} \geq \frac{0.010184}{0.006} = \frac{0.10092}{0.006} \approx 16.8 rad/s .$ $16.8$ rad/s se neeche, koi steady precession possible nahi.

L3.2 L3.1 ke top ko exactly minimum spin par spin kiya jaaye, to (single, double) precession rate $Ω$ kya hai?

Recall Solution

KYA: minimum spin par discriminant zero hota hai, isliye dono roots ek mein collapse ho jaate hain. KYUN: $0 = 0$ se $Ω = I_{3} ω_{s} / (2 I_{1} cos θ)$ bachta hai. $ω_{s} = 16.82$ rad/s ke saath: $Ω = \frac{I _{3} ω _{s}}{2 I _{1} c o s θ} = \frac{0.006 \cdot 16.82}{2 \cdot 0.012 \cdot 0.8660} = \frac{0.10092}{0.020785} \approx 4.86 rad/s .$ Fast aur slow precessions yahan ek single "borderline" rate mein merge ho jaati hain.

L3.3 Algebraically dikhao ki fast-top formula large-spin limit mein exact quadratic ka $-$ root hai. L3.1 top ko $ω_{s} = 100$ rad/s par spin karne par numerically verify karo.

Recall Solution

KYA: large $ω_{s}$ ke liye square root expand karo. KYUN: hum dekhna chahte hain ki exact theory simple theory ko contain karti hai. $a = I_{3} ω_{s}$ , $b = 4 I_{1} m g ℓ cos θ$ ke saath: $a^{2} - b \approx a - \frac{b}{2 a}$ , isliye $Ω_{-} \approx \frac{a - ( a - b /2 a )}{2 I _{1} c o s θ} = \frac{b /2 a}{2 I _{1} c o s θ} = \frac{m g ℓ}{I _{3} ω _{s}} .$ Numeric check $ω_{s} = 100$ par: fast-top deta hai $Ω = (0.5 \cdot 9.8 \cdot 0.05) / (0.006 \cdot 100) = 0.2450/0.6 = 0.4083$ rad/s. Exact $-$ root: $a = 0.6$ , $b = 0.010184$ , $a^{2} - b = 0.36 - 0.010184 = 0.349816 = 0.59145$ ; $Ω_{-} = (0.6 - 0.59145) / (2 \cdot 0.012 \cdot 0.8660) = 0.008554/0.020785 = 0.4116$ rad/s. Agreement lagbhag $0.8%$ tak. ✓

Level 4 — Synthesis

L4.1 Ek gyroscope $Ω_{1} = 0.98$ rad/s par precess karta hai (L1 wheel). Ab tum spin $ω_{s}$ ko touch kiye bina precession rate double karna chahte ho. Do alag single-parameter changes batao jo yeh achieve karein, aur har ek ki nai value bolo.

Recall Solution

KYA: padho ki $Ω$ ko kaun se factors control karte hain aur ek ko scale karo. KYUN: $Ω = m g ℓ / (I_{3} ω_{s})$ , $m$ , $ℓ$ , aur $1/ I_{3}$ mein linear hai. Fixed $ω_{s}$ par $Ω$ double karne ke liye, inme se koi ek:

$ℓ$ double karo: $ℓ = 0.40$ m ⟹ $Ω = 3.92 \cdot 2/4.0 = 1.96$ rad/s.
$m$ double karo: $m = 4.0$ kg ⟹ $Ω = 2 \cdot 3.92/4.0 = 1.96$ rad/s.
$I_{3}$ aadha karo: $I_{3} = 0.05 kg\cdotm^{2}$ ⟹ $Ω = 3.92/ (0.05 \cdot 40) = 1.96$ rad/s. Teeno mein se koi bhi kaam karta hai; har ek $Ω = 1.96$ rad/s deta hai.

L4.2 Ek wheel ko spin up kiya jaata hai taaki uska spin angular momentum $L_{s} = I_{3} ω_{s} = 5.0 kg\cdotm^{2} / s$ ho, tilt $θ = 6 0^{\circ}$ par, gravitational torque magnitude $∣ τ ∣ = m g ℓ sin θ = 3.0$ N·m ke saath. Seedha $∣ τ ∣ = Ω L_{s} sin θ$ use karke (parent ka Step 2) $Ω$ nikalo, aur confirm karo ki yeh $m g ℓ / (I_{3} ω_{s})$ ke barabar hai.

Recall Solution

KYA: $Ω$ banane ke liye pre-cancellation torque balance use karo, phir check karo ki $sin θ$ sach mein cancel hota hai. KYUN: yeh "remarkable cancellation" ko symbolic ki jagah concrete banata hai. Pehle $m g ℓ$ extract karo: kyunki $m g ℓ sin θ = 3.0$ aur $sin 6 0^{\circ} = 0.8660$ , hume $m g ℓ = 3.0/0.8660 = 3.4641$ N·m milta hai. Balance: $Ω L_{s} sin θ = m g ℓ sin θ$ , isliye $Ω = \frac{m g ℓ s i n θ}{L _{s} s i n θ} = \frac{m g ℓ}{L _{s}} = \frac{3.4641}{5.0} = 0.6928 rad/s .$ Boxed formula se cross-check: $m g ℓ / (I_{3} ω_{s}) = 3.4641/5.0 = 0.6928$ rad/s. ✓ Dono $sin θ$ exactly cancel ho gaye.

L4.3 Ideas combine karo: ek top jisme $I_{3} = 0.006$ , $I_{1} = 0.012 kg\cdotm^{2}$ , $m = 0.5$ kg, $ℓ = 0.05$ m, $θ = 4 5^{\circ}$ par $ω_{s} = 30$ rad/s spin karta hai. Dono exact precession rates $Ω_{+}$ (fast) aur $Ω_{-}$ (slow) nikalo, aur identify karo ki fast-top formula kaun sa approximate karta hai.

Recall Solution

KYA: full quadratic ke dono roots evaluate karo. KYUN: exact theory ek fast (nutation-like) branch aur ek slow (gravity-driven) branch predict karta hai — derivation ke liye Lagrangian Mechanics dekho. Ingredients: $a = I_{3} ω_{s} = 0.006 \cdot 30 = 0.18$ ; $b = 4 I_{1} m g ℓ cos 4 5^{\circ} = 4 \cdot 0.012 \cdot 0.5 \cdot 9.8 \cdot 0.05 \cdot 0.7071 = 0.0083160$ . Discriminant: $a^{2} - b = 0.0324 - 0.0083160 = 0.0240840$ ; $= 0.155190$ . Denominator: $2 I_{1} cos 4 5^{\circ} = 2 \cdot 0.012 \cdot 0.7071 = 0.016971$ . $Ω_{+} = \frac{0.18 + 0.155190}{0.016971} = \frac{0.335190}{0.016971} = 19.75 rad/s (fast),$ $Ω_{-} = \frac{0.18 - 0.155190}{0.016971} = \frac{0.024810}{0.016971} = 1.462 rad/s (slow) .$ Fast-top formula slow root $Ω_{-}$ ko approximate karta hai: check karo $m g ℓ / (I_{3} ω_{s}) = (0.5 \cdot 9.8 \cdot 0.05) /0.18 = 0.245/0.18 = 1.361$ rad/s — $1.462$ ke kareeb. Dono roots yahan positive hain (prograde) kyunki $cos 4 5^{\circ} > 0$ .

Level 5 — Mastery

L5.1 Quadratic $I_{1} cos θ Ω^{2} - I_{3} ω_{s} Ω + m g ℓ = 0$ se directly dono exact roots ka sum aur product derive karo, phir unhe L4.3 numbers bina re-solving ke verify karne ke liye use karo.

Recall Solution

KYA: Vieta's relations apply karo. KYUN: $A Ω^{2} + B Ω + C = 0$ ke liye, roots satisfy karte hain $Ω_{+} + Ω_{-} = - B / A$ aur $Ω_{+} Ω_{-} = C / A$ — ek fast cross-check. Yahan $A = I_{1} cos θ$ , $B = - I_{3} ω_{s}$ , $C = m g ℓ$ . Isliye: $Ω_{+} + Ω_{-} = \frac{I _{3} ω _{s}}{I _{1} c o s θ}, Ω_{+} Ω_{-} = \frac{m g ℓ}{I _{1} c o s θ} .$ L4.3 sum check: $\frac{I _{3} ω _{s}}{I _{1} cos θ} = \frac{0.18}{0.012 \cdot 0.7071} = \frac{0.18}{0.0084853} = 21.21$ . Direct: $19.75 + 1.462 = 21.21$ . ✓ L4.3 product check: $\frac{m g ℓ}{I _{1} cos θ} = \frac{0.245}{0.0084853} = 28.87$ . Direct: $19.75 \cdot 1.462 = 28.88$ . ✓

L5.2 Algebraically dikhao ki slow-spin limit $ω_{s} \to 0$ mein (ek non-spinning pivoted rod jo angle $θ$ par release hui ho), product-of-roots result force karta hai ki dono precession rates magnitude mein equal aur sign mein opposite hon. Physically interpret karo.

Recall Solution

KYA: sum aur product mein $ω_{s} = 0$ set karo. KYUN: sum $Ω_{+} + Ω_{-} = I_{3} ω_{s} / (I_{1} cos θ) \to 0$ , isliye $Ω_{-} = - Ω_{+}$ . Product $Ω_{+} Ω_{-} = m g ℓ / (I_{1} cos θ) > 0$ for $θ < 9 0^{\circ}$ — lekin agar $Ω_{-} = - Ω_{+}$ to product $- Ω_{+}^{2} < 0$ hai. Contradiction ⟹ non-spinning top ke liye koi real steady precession exist nahi karti (discriminant $- 4 I_{1} m g ℓ cos θ < 0$ ). Interpretation: spin ke bina koi gyroscopic stiffness nahi hoti; ek released rod bas girta hai (yeh pure Nutation amplitude hai, steady precession nahi). Steady precession fundamentally ek spin phenomenon hai.

L5.3 Capstone. Ek gyroscope demonstration ko exactly $T = 5.0$ s period ke saath $θ = 3 0^{\circ}$ tilt par precess karna hai, $m = 1.0$ kg, $ℓ = 0.15$ m, $I_{1} = 0.020 kg\cdotm^{2}$ , $I_{3} = 0.010 kg\cdotm^{2}$ ke saath. Exact slow root use karke required spin $ω_{s}$ nikalo. (Hint: $T$ ko $Ω$ mein badlo, phir quadratic ko $ω_{s}$ ke liye solve karo.)

Recall Solution

KYA: $Ω = 2 π / T$ target karo, phir quadratic rearrange karke $ω_{s}$ isolate karo. KYUN: equation $Ω$ mein quadratic hai lekin $ω_{s}$ sirf ek single $- I_{3} ω_{s} Ω$ term mein aata hai, isliye yeh $ω_{s}$ mein linear hai aur cleanly solve ho jaata hai. Step 1 — target rate: $Ω = 2 π / T = 2 π /5.0 = 1.2566$ rad/s. Step 2 — $ω_{s}$ isolate karo: $I_{1} Ω^{2} cos θ - I_{3} ω_{s} Ω + m g ℓ = 0$ se shuru karo. $ω_{s}$ term akele ek taraf le jao: $I_{3} ω_{s} Ω = I_{1} Ω^{2} cos θ + m g ℓ ⟹ ω_{s} = \frac{I _{1} Ω ^{2} c o s θ + m g ℓ}{I _{3} Ω} .$ Step 3 — numbers plug karo. Numerator: $I_{1} Ω^{2} cos θ = 0.020 \cdot (1.2566)^{2} \cdot 0.8660 = 0.020 \cdot 1.5790 \cdot 0.8660 = 0.027348$ ; add $m g ℓ = 1.0 \cdot 9.8 \cdot 0.15 = 1.470$ ; total $= 1.497348$ . Denominator: $I_{3} Ω = 0.010 \cdot 1.2566 = 0.0125664$ . $ω_{s} = \frac{1.497348}{0.0125664} = 119.2 rad/s$ Step 4 — fast-top se sanity check: $m g ℓ / (I_{3} Ω) = 1.470/0.0125664 = 116.98$ rad/s — exact answer thoda zyada hai kyunki small centrifugal term $I_{1} Ω^{2} cos θ$ numerator mein add hota hai. ✓

L5.4 Edge case. Horizontal axle, $θ = 9 0^{\circ}$ (isliye $cos θ = 0$ ) ke liye reasoning repeat karo. Exact-root formula mein kya gadbad hoti hai, aur wahaan sahi steady precession rate kya hai?

Recall Solution

KYA: $cos 9 0^{\circ} = 0$ substitute karo aur exact formula ka denominator vanish hote dekho. KYUN: $2 I_{1} cos θ \to 0$ se $Ω_{\pm} = \frac{I _{3} ω _{s} \pm ( I _{3} ω _{s} ) ^{2} - 0}{0}$ — ek $\frac{something}{0}$ , isliye quadratic degenerate ho jaata hai. Divide karne se pehle equation par wapas jao, $cos θ = 0$ par: $I_{1} Ω^{2} \cdot 0 - I_{3} ω_{s} Ω + m g ℓ = 0 ⟹ - I_{3} ω_{s} Ω + m g ℓ = 0.$ $Ω^{2}$ term chala gaya: quadratic ek linear equation mein collapse ho jaata hai jisme ek finite root hai $Ω = \frac{m g ℓ}{I _{3} ω _{s}},$ jo exactly fast-top formula hai — $θ = 9 0^{\circ}$ par koi approximation nahi chahiye. "Second root" $Ω_{+}$ $\pm \infty$ tak chala gaya; physically fast nutation-scale branch disappear ho jaati hai kyunki spin axis ka koi vertical component nahi hai jisme transverse inertia $I_{1}$ precession angular momentum store kar sake. Interpretation: ek horizontal gyroscope (axle level, CM sideways offset) precisely $m g ℓ / (I_{3} ω_{s})$ par precess karta hai bina kisi correction ke. $θ$ ke $9 0^{\circ}$ se thoda neeche ke liye, $Ω_{-}$ finite rehta hai aur is value ke kareeb hota hai jabki $Ω_{+}$ blow up karta hai; $θ > 9 0^{\circ}$ ( $cos θ < 0$ ) cross karne par dono roots ka sign flip hota hai → retrograde precession, upar wale sign convention ke consistent.

Recall Self-test cloze summary

Fast-top precession rate ::: $Ω = \frac{m g ℓ}{I _{3} ω _{s}}$ , tilt $θ$ se independent. Kaun sa moment of inertia spin angular momentum carry karta hai ::: $I_{3}$ (symmetry axis ke baare mein). Real steady precession ke liye condition ::: $(I_{3} ω_{s})^{2} \geq 4 I_{1} m g ℓ cos θ$ . Fast-top formula kaun sa exact root approximate karta hai ::: slow root $Ω_{-}$ . Negative $Ω$ ka kya matlab hai ::: opposite (retrograde/clockwise-from-above) sense mein precession, koi error nahi. $θ = 9 0^{\circ}$ par exact formula ka kya hota hai ::: denominator $2 I_{1} cos θ \to 0$ ; quadratic linear mein degenerate ho jaata hai aur $Ω = m g ℓ / (I_{3} ω_{s})$ exactly. Spin khatam hone par precession ka kya hota hai ::: speed up hoti hai ( $Ω \propto 1/ ω_{s}$ ), phir nutate karti hai aur girta hai.