Question bank — Angular momentum L = Iω (fixed axis), L = r × p (general)

False — $L$ sirf us particle ki line of motion par ke points ke baare mein zero hota hai; ek off-line origin ke baare mein lever arm $r_{\perp }\ne 0$ hota hai, isliye $L=p{r}_{\perp }$ nonzero hai aur constant rehta hai.

False — ek body jo $\omega =0$ ke saath slide (translate) kar rahi hai wo phir bhi apne path se door kisi bhi origin ke baare mein orbital angular momentum $\vec{L}=\vec{r}\times \vec{p}$ carry karti hai; "not spinning" aur "zero $L$" alag alag cheezein hain.

False — $\vec{L}=\vec{r}\times \vec{p}$ chosen origin par depend karta hai; origin badlo aur $\vec{r}$ aur $\vec{L}$ dono badal jaate hain. Koi "the" angular momentum exist nahi karta jab tak ek point naam na lo.

False — $L$ perpendicular distance $r_{\perp }$ (lever arm) aur motion ki direction par bhi depend karta hai; same $m$ aur $v$ wildly different $L$ de sakte hain.

False — ye sirf ek principal/symmetry axis ke baare mein hold karta hai. Generally $I$ ek tensor hai aur $\vec{L}$ spin axis se tilt ho sakta hai, yehi exact reason hai ki ek unbalanced wheel wobble karta hai (neeche ka tilt figure dekho).

False — skater jo arms andar kheenchti hai wo $L=I\omega$ fixed rakhti hai lekin $KE=\frac{1}{2}I{\omega }^2=L^2/(2I)$ badhti hai jab $I$ ghatta hai; uski muscles extra energy supply karti hain. $L$ conserved hone ka matlab $KE$ conserved hona nahi hai.

False — $L$ ki units $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}=\mathrm{J}\mathrm{s}$ hain (energy × time), joules nahi. Time ka extra factor hi giveaway hai.

False — sirf torque $\vec{\tau }=\vec{r}\times \vec{F}$ hi $\vec{L}$ ko change karta hai. Origin ki taraf ya us se seedha door point karne wali force (central force) mein $\vec{r}\parallel \vec{F}$ hota hai, isliye $\vec{\tau }=0$ aur $L$ conserved rehta hai — yehi exact reason hai ki planets Kepler's Second Law maan'te hain.

False — $I=\sum m_i{r}_i^2$ chosen axis par depend karta hai; same disc ka $I=\frac{1}{2}M{R}^2$ uske centre ke baare mein hai lekin edge ke baare mein ek bada value hai (parallel-axis theorem).

Sahi magnitude $L=rp\sin \theta =p{r}_{\perp }$ hai. $\sin \theta$ chhod dena ignore karna hai ki sirf motion ka wo component origin ke around jaata count hota hai; radial part kuch contribute nahi karta.

Galat scope — $\vec{r}\times \vec{p}$ per particle hota hai. Poore rigid body ke liye tumhe saare particles par $m_i{r}_i^2$ sum karna hoga, jo $L=I\omega$ deta hai; centre ka apna $\vec{r}\times \vec{p}$ internal spin ko bilkul miss kar deta hai.

Sahi condition zero external torque hai, zero force nahi. Ek force act kar sakti hai lekin koi torque produce na kare (central force), ya torque exist kar sakta hai balanced forces ke saath (a couple). $L$ ka conservation ${\vec{\tau }}_{ext}=0$ se tied hai.

Koi external torque act nahi karta; uski apni muscles internal forces hain jo $I$ reduce karti hain. $L=I\omega$ fixed rakhne ke liye, $\omega$ badhna chahiye — speed-up ek bookkeeping consequence hai, external push nahi.

Circular motion ke liye ye actually theek hai — velocity radius ke perpendicular hoti hai isliye $\sin \theta =1$. Error ye hai ki us shortcut ko non-circular motion par generalise karna, jahan $\sin \theta \ne 1$ aur $r_{\perp }$ use karna zaroori hai.

Ye alag alag quantities hain: $\vec{\tau }=d\vec{L}/dt$ $\vec{L}$ ka rate of change hai, jaise force linear momentum ka rate of change hai. Constant $\vec{L}$ zero $\vec{\tau }$ ke saath coexist kar sakta hai; nonzero $\vec{\tau }$ $\vec{L}$ ke sideways point kar sakta hai aur use sirf turn kar sakta hai.

Cross product $\vec{r}\times \vec{p}$ automatically motion ko discard karta hai jo origin ki taraf seedha point karta hai (wo koi turning produce nahi karta) aur sirf "swirl" component rakhta hai — exactly wo part jo origin ke baare mein rotation measure karta hai. Cross Product aur figure s01 dekho.

Kyunki $I$ mass ka rotational analogue hai — ye resistance to changing spin measure karta hai. Lekin mass ke unlike, $I$ depend karta hai ki mass axis se kitna door baitha hai ($I=\sum m_i{r}_i^2$), isliye mass ko andar move karna isse change karta hai. Warning: $I$ ek scalar hai sirf ek symmetry axis ke baare mein; off-axis (figure s02) har particle ka ${\vec{r}}_i\times {\vec{p}}_i$ thoda tilt hota hai, aur tilts add up hote hain taaki $\vec{L}$ $\vec{\omega }$ se off point kare.

Koi force nahi hai toh koi torque nahi, isliye $\vec{L}$ change nahi ho sakta. Geometrically (figure s03): ball ka seedha path ek fixed line hai, aur fixed origin se fixed line tak ki perpendicular distance ek single unchanging number hai — wo hai $r_{\perp }$. Isliye jab $\vec{r}$ lengthens aur rotate karta hai, path par uska perpendicular drop sama rehta hai, aur $p$ constant hai (koi force nahi), $L=p{r}_{\perp }$ fixed rakhta hai.

Kyunki $\vec{L}=\vec{r}\times \vec{p}$ ke liye position vector $\vec{r}$ chahiye, aur "position" meaningless hai jab tak tum wo point fix na karo jahan se measure ho raha hai. Different origins different lever arms aur isliye different $L$ dete hain.

Gravity ek central force hai jo $\vec{r}$ ke along Sun ki taraf point karti hai, isliye $\vec{\tau }=\vec{r}\times \vec{F}=0$ aur $\vec{L}$ conserved hai. Geometrically (figure s03): ek chote time $dt$ mein planet $\vec{v}dt$ move karta hai, ek patla triangle sweep karta hai jiska area $dA=\frac{1}{2}|\vec{r}\times \vec{v}dt|=\frac{1}{2m}|\vec{r}\times \vec{p}|dt=\frac{L}{2m}dt$ hai. Isliye $dA/dt=L/(2m)$, ek constant — equal areas in equal times.

$I$ poori body ki ek axis ke baare mein property hai; ek single particle sirf $m_i{r}_i^2$ contribute karta hai. Moment of inertia tabhi meaningful hota hai jab har particle ka contribution add up ho jaata hai. Moment of Inertia dekho.

Exactly zero — $\vec{r}$ aur $\vec{p}$ antiparallel hain, isliye $\sin \theta =0$ aur $\vec{r}\times \vec{p}=\vec{0}$. Saari motion radial hai; kuch bhi "around nahi jaata."

Zero, kyunki $\vec{r}=\vec{0}$ se $\vec{r}\times \vec{p}=\vec{0}$ milta hai chahe wo kitni bhi tezi se move kare. Ek body ka us point ke baare mein koi lever arm nahi hota jis par wo occupy karta hai.

Net force zero hai lekin net torque nahi, isliye $d\vec{L}/dt\ne 0$ aur $\vec{L}$ change hota hai. Ye clean case prove karta hai ki conservation torque par depend karta hai, force par nahi.

Haan, aur ye $L=0$ us axis ke baare mein deta hai — ek degenerate lekin consistent case. Formula hold karta hai; ye sirf report karta hai ki ek non-spinning rigid body koi spin angular momentum carry nahi karta.

Isse split karo: ${\vec{L}}_O={\vec{r}}_{cm}\times {\vec{p}}_{cm}+{\vec{L}}_{spin}$. Pehla term orbital part hai (centre of mass origin ke past move kar raha hai, jaise saari mass wahan baithti ho), doosra spin part hai ($I_{cm}\omega$) centre ke baare mein. Figure s04 dono pieces ko stack hote dikhata hai jo total deta hai.

$\vec{L}=I\vec{\omega }$ bhi sign flip kar leta hai, axis ke along opposite direction mein point karta hai, kyunki $I>0$ ek positive scalar hai. Magnitude unchanged rehta hai; sirf right-hand-rule direction reverse hoti hai.

Nahi — $\vec{L}$ axis se tilt ho jaata hai (figure s02), aur uska component jo $\vec{\omega }$ ke perpendicular hai sweep karta hai, bearings ko har turn mein ek torque supply karne par majboor karta hai. Sirf axial component simple $I\omega$ jaisa behave karta hai.