Exercises — Gyroscopic effect — precession of spinning top

(a) $\vec{L}$ spin axis ke saath point karta hai (upar aur tilt hua; sense spin ke around right-hand rule se set hota hai). (b) $\vec{\tau }=\vec{r}\times m\vec{g}$ horizontal hai, us vertical plane ke perpendicular jisme axis hai. Gravity seedha neeche hai, $\vec{r}$ axis ke upar hai, toh unka cross product horizontal hai — yeh kabhi neeche point nahi kar sakta. (c) $d\vec{L}=\vec{\tau }dt$ $\vec{\tau }$ ke same direction mein point karta hai — horizontal. Isliye axis sideways swing karta hai (precess karta hai) girne ki bajaye. Neeche in teeno vectors ki picture dekho.

Upar (faster precession): arm length $r$, mass $m$, gravity $g$ — yeh sab milke torque $rmg$ hain. Neeche (slower precession): moment of inertia $I$ aur spin rate $\omega$ — yeh sab milke spin angular momentum $L=I\omega$ hain. Tilt $\theta$: yeh cancel ho gaya. Yeh ek baar $\tau =rmg\sin \theta$ mein appear karta hai (numerator) aur ek baar horizontal projection $L\sin \theta$ mein (denominator), toh $\Omega$ fast-spin limit mein tilt se independent hai.

Step 1 — torque. $\tau =rmg=0.10\times 2.0\times 9.8=1.96\text{N⋅m}$. Step 2 — angular momentum. $L=I\omega =0.050\times 150=7.5{\text{kg⋅m}}^2/\text{s}$. Step 3 — precession rate. $\Omega =\tau /L=1.96/7.5=0.2613\text{rad/s}$. Step 4 — period. $T=2\pi /\Omega =6.2832/0.2613=24.0\text{s}$. Axis har 24 seconds mein roughly ek poora horizontal circle sweep karta hai.

Step 1. $I=m{R}^2=3.0\times (0.35{)}^2=3.0\times 0.1225=0.3675{\text{kg⋅m}}^2$. Step 2. $L=I\omega =0.3675\times 25=9.1875{\text{kg⋅m}}^2/\text{s}$. Step 3. $\tau =rmg=0.25\times 3.0\times 9.8=7.35\text{N⋅m}$. Step 4. $\Omega =\tau /L=7.35/9.1875=0.800\text{rad/s}$.

$\Omega =rmg/(I\omega )$, toh $\Omega \propto 1/\omega$. $\omega$ ko triple karne se $\Omega$ ek-tihai reh jaata hai. Period rate ka inverse hai: $T=2\pi /\Omega$, toh $T\propto \omega$. $\omega$ ko triple karne se period triple ho jaata hai. $T_2=3\times 6.0=18.0\text{s}$.

Kisi bhi $\theta$ par: $\tau =rmg\sin \theta =0.50\sin \theta$, aur horizontal projection $L\sin \theta =2.0\sin \theta$ hai. $\theta =30°$: $\Omega =\frac{0.50\sin 30°}{2.0\sin 30°}=\frac{0.50}{2.0}=0.25\text{rad/s}$. $\theta =75°$: $\Omega =\frac{0.50\sin 75°}{2.0\sin 75°}=\frac{0.50}{2.0}=0.25\text{rad/s}$. $\sin \theta$ upar aur neeche dono se divide ho jaata hai — dono tilts ke liye same $\Omega =0.25\text{rad/s}$. Neeche ka figure dikhata hai kyun: $\vec{L}$ ki tip ke trace kiye horizontal circle ka radius $L\sin \theta$ hai, jo bilkul match karta hai ki torque kaise badhta hai.

$\vec{r}=r\hat{x}$ likho (iska horizontal lean; vertical part $\vec{g}$ ke parallel hai aur cross product mein kuch contribute nahi karta), aur $m\vec{g}=-mg\hat{z}$. (a) $\vec{\tau }=(r\hat{x})\times (-mg\hat{z})=-rmg(\hat{x}\times \hat{z})$. Ab $\hat{x}\times \hat{z}=-\hat{y}$, toh $\vec{\tau }=-rmg(-\hat{y})=+rmg\hat{y}$. $\vec{\tau }$ $+y$ direction mein point karta hai. (b) $d\vec{L}=\vec{\tau }dt$ $+y$ ke saath hai, toh axis (jo $+x$ ki taraf jhuki thi) $+y$ ki taraf swing karegi. (c) $+x\to +y\to -x\to -y$ jaana upar se dekha jaaye toh counter-clockwise hai ($+z$ axis ke upar se neeche dekhte hue). Figure dekho.

Step 1 — required $\Omega$. $\Omega =2\pi /T=6.2832/10.0=0.62832\text{rad/s}$. Step 2 — precession formula invert karo. $\Omega =rmg/(I\omega )\Rightarrow \omega =\frac{rmg}{I\Omega }$. Step 3 — torque. $rmg=0.12\times 1.5\times 9.8=1.764\text{N⋅m}$. Step 4 — solve karo. $\omega =\frac{1.764}{0.020\times 0.62832}=\frac{1.764}{0.0125664}=140.4\text{rad/s}$.

(a) $\tau =rmg=0.03\times 0.20\times 9.8=0.0588\text{N⋅m}$. $L=I\omega =0.004\times 8.0=0.032{\text{kg⋅m}}^2/\text{s}$. $\Omega =\tau /L=0.0588/0.032=1.8375\text{rad/s}$. (b) Ratio $\Omega /\omega =1.8375/8.0=0.2297$, yaani lagbhag $23\%$. Yeh $\ll 1$ nahi hai. Fast-spin approximation shaky hai: jo physics neglect ho rahi hai woh nutation hai — axis precess karte waqt clearly upar-neeche bob karega, aur sach mein averaged $\Omega$ is value se thoda alag hoga. Clean steady precession ke liye $\omega$ ko das ya zyada times bada chahiye hoga.

(a) Jaise $\omega \to 0$, denominator $I\omega \to 0$, toh $\Omega \to \infty$. Formula infinitely fast precession predict karta hai. (b) Physically ek non-spinning top simply gir jaata hai — koi precession nahi. Yeh contradiction formula ka apna breakdown dikhana hai: $\omega =0$ ke saath koi angular momentum redirect karne ke liye nahi hai, toh $\vec{\tau }=d\vec{L}/dt$ simply $\vec{L}$ ko torque ki direction mein zero se build up karta hai — top topple kar jaata hai. Formula $\Omega =rmg/(I\omega )$ sirf fast-spin limit mein valid hai jahan pehle se ek bada $\vec{L}$ exist karta hai jo steer kiya ja sake; "$\Omega \to \infty$" iska tarika hai yeh kehne ka ki "main yahan apply nahi hota."

(a) ${\Omega }_A=\tau /L_A=2.0/4.0=0.50\text{rad/s}$. ${\Omega }_B=\tau /L_B=2.0/12.0=0.1\overline{6}=0.1667\text{rad/s}$. (b) ${\Omega }_A/{\Omega }_B=0.50/0.1667=3.0$. Gyro B precesses 3× slower — bilkul ratio $L_B/L_A=3$ jaisa. Bada $L$ ⇒ same torque par slower response. (c) Ek bullet jise bada spin $L$ diya gaya ho, uska kisi bhi cross-wind ya gravity torque ke liye tiny $\Omega$ hoga, toh flight ke dauran uski naak barely axis se bhatakti hai — yahi spin-stabilisation hai, wahi principle jo Gyroscope and Navigation system mein use hoti hai.

• Precession rate formula? ::: $\Omega =\tau /(L\sin \theta )=rmg/(I\omega )$.
• $d\vec{L}$ ki direction? ::: $\vec{\tau }=\vec{r}\times m\vec{g}$ ke same (horizontal).
• Spin triple karo ⇒ period kya karta hai? ::: Triple ho jaata hai ($T\propto \omega$).
• $\sin \theta$ cancel kyun hota hai? ::: Yeh torque aur horizontal $L$-projection dono ko multiply karta hai.
• Formula kab fail karta hai? ::: Jab $\omega$, $\Omega$ se $\gg$ nahi hota (nutation matter karta hai); $\omega =0$ par yeh galat $\infty$ deta hai.