Visual walkthrough — Rotating frames — centrifugal force, Coriolis force

1.2.14 · D2 · Physics › Newton's Laws & Dynamics › Rotating frames — centrifugal force, Coriolis force

Hum sirf yeh assume karte hain ki tum:

• arrows ko tip-to-tail jod sakte ho (vectors),
• aur ek ghoomta hua merry-go-round imagine kar sakte ho.

Baaki sab — including cross product — hum jaisa zaroorat padega waisa banate jayenge.

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Step 1 — Do observers, ek arrow

KYA. Ek flat spinning disc (merry-go-round) imagine karo. Ek observer upar float kar rahi hai, spinning nahi — use inertial observer bulao. Doosri disc pe baith ke, uske saath ghoom rahi hai — rotating observer. Disc pe hum do arrows paint karte hain, ${\hat{e}}_1$ aur ${\hat{e}}_2$, floor se chipke hue. Yeh hamare basis vectors hain: length 1 ke chhote rulers jo batate hain "yeh taraf x hai, woh taraf y hai."

KYUN. Har position, velocity, ya force sirf inn do rulers ka combination hai. Agar hum samjhein ki rulers khud kaise chalte hain, toh hum sab kuch samajh lenge. Pakda yeh hai: rotating observer ke liye rulers still baithe hain; inertial observer ke liye rulers ghoomte hain. Yahi disagreement poora subject hai.

PICTURE. Figure mein, black arrows ${\hat{e}}_1,{\hat{e}}_2$ disc ke saath point karte hain. Curved gray arrow dikhata hai disc rate $\omega$ (radians per second) pe ghoom rahi hai. Floating observer (upar) arrows ko rotate hote dekhti hai; baithi observer (disc pe) unhe frozen dekhti hai.

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Step 2 — Ek chipka hua ruler kaise chalta hai: cross product appear hota hai

KYA. Ek ruler ki tip, ${\hat{e}}_1$, ko thodi si time mein dekho. Woh stretch nahi hoti (abhi bhi length 1); woh sirf sideways swing karti hai. Hum us tip ki sideways velocity ke liye ek formula chahte hain.

KYUN yeh tool — cross product. Hume ek aise operation ki zaroorat hai jo spin arrow $\vec{\omega }$ aur ruler ${\hat{e}}_1$ leke ek teesra arrow perpendicular dono se return kare, jiska length "spin rate × tip ki axis se door kitni hai" ho. Yahi bilkul cross product $\vec{\omega }\times {\hat{e}}_1$ karta hai — koi aur simple product perpendicular direction nahi deta. Hum ise isliye choose karte hain kyunki tip ko perpendicular chalna hai dono axis aur ruler ke (woh axis ke around circle karti hai).

PICTURE. ${\hat{e}}_1$ ki tip ek circle trace karti hai. Green arrow $\vec{\omega }\times {\hat{e}}_1$ us circle ka tangent hai — yeh tip ki instantaneous velocity hai. Lamba $\vec{\omega }$ (tezi se spin) ⇒ lamba green arrow.

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Step 3 — Transport Theorem: do observers ke beech translate karna

KYA. Koi bhi arrow $\vec{A}$ lo (position, velocity, kuch bhi). Hume koi bhi arrow likhne ke liye rulers ka poora set of three chahiye, toh in-disc rulers ${\hat{e}}_1,{\hat{e}}_2$ ke saath ek teesra add karte hain:

Teeno rulers ke saath, kisi bhi arrow $\vec{A}$ ke teen numbers hote hain: $\vec{A}=A_1{\hat{e}}_1+A_2{\hat{e}}_2+A_3{\hat{e}}_3.$ Ab poochho ki $\vec{A}$ inertial observer ko kitni fast change hoti hai.

KYUN. Ek arrow do independent reasons se change ho sakti hai: (1) iske numbers $A_1,A_2,A_3$ change hote hain — rotating observer yeh dekhti hai — ya (2) uske neeche ke rulers ghoom jaate hain — sirf inertial observer yeh dekhti hai. Dono add karne chahiye. Yeh sirf differentiation ka product rule hai, "number × ruler" pe apply kiya.

PICTURE. Figure ek blue arrow ki motion ko do red pieces mein todta hai: ek seedha piece (numbers change ho rahe hain, disc pe dikh raha hai) plus ek curved sideways piece (rulers rotate ho rahe hain). Unka tip-to-tail sum woh hai jo floating observer dekhti hai.

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Step 4 — Pehla application: velocities

KYA. Position arrow $\vec{r}$ (axis se object tak) ko transport theorem mein feed karo.

KYUN. Velocity hoti hi hai position ke change ki rate. Toh $\vec{A}=\vec{r}$ plug karna hume batata hai ki do observers ki velocities kaise relate hoti hain — acceleration tak seedhi ladder ki pehli seedi.

PICTURE. Ek person outward chalta hai (orange ${\vec{v}}_{\mathrm{rot}}$) jab floor use sideways sweep karta hai (green $\vec{\omega }\times \vec{r}$). Sach mein inertial velocity (blue) diagonal sum hai.

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Step 5 — Doosra application: acceleration (jahan do forces paida hoti hain)

KYA. Operator ko phir apply karo, is baar ${\vec{v}}_{\mathrm{in}}={\vec{v}}_{\mathrm{rot}}+\vec{\omega }\times \vec{r}$ par. Hum $\vec{\omega }$ constant rakhte hain (steady spin).

KYUN. Acceleration velocity ke change ki rate hai, aur Newton sirf acceleration ki language mein bolta hai ($\vec{F}=m\vec{a}$, dekho Newton's Second Law). Toh forces ki language tak pahunchne ke liye ek baar aur differentiate karna padega.

${\left(\frac{d}{dt}\right)}_{\mathrm{in}}={\left(\frac{d}{dt}\right)}_{\mathrm{rot}}+\vec{\omega }\times$ ko har piece pe apply karte hain:

${\vec{a}}_{\mathrm{in}}=\underset{={\vec{a}}_{\mathrm{rot}}}{\underbrace{(\frac{d}{dt}{)}_{\mathrm{rot}}{\vec{v}}_{\mathrm{rot}}}}+\underset{\text{from operator on }{\vec{v}}_{\mathrm{rot}}}{\underbrace{\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}}}+\underset{=\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}}{\underbrace{(\frac{d}{dt}{)}_{\mathrm{rot}}(\vec{\omega }\times \vec{r})}}+\underset{\text{from operator on }\vec{\omega }\times \vec{r}}{\underbrace{\vec{\omega }\times (\vec{\omega }\times \vec{r})}}$

Dekho 2 ka factor kaise appear hota hai. Do alag terms ne $\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$ diya: ek $\vec{\omega }\times$ operator se jo ${\vec{v}}_{\mathrm{rot}}$ hit karta hai, ek rotating frame ke andar $\vec{\omega }\times \vec{r}$ differentiate karne se (kyunki $(d\vec{r}/dt{)}_{\mathrm{rot}}={\vec{v}}_{\mathrm{rot}}$). Woh add ho jaate hain:

PICTURE. Figure track karta hai $\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$ ke do clones ko jo ek doubled green arrow mein merge hote hain — isliye 2 wahan hai, typo nahi hai.

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Step 6 — Newton insert karo, fictitious forces padho

KYA. Real forces obey karte hain ${\vec{F}}_{\mathrm{real}}=m{\vec{a}}_{\mathrm{in}}$. Boxed line substitute karo aur solve karo ki baithi observer ko kya chahiye, $m{\vec{a}}_{\mathrm{rot}}$.

KYUN. Baithi observer simple rule use karna chahti hai "force = mass × woh acceleration jo main measure karti hun." Hum rearrange karte hain taaki leftover $\vec{\omega }$-terms force side pe aa jayein aur forces ki tarah masquerade karein.

$m{\vec{a}}_{\mathrm{rot}}=m{\vec{a}}_{\mathrm{in}}-2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}-m\vec{\omega }\times (\vec{\omega }\times \vec{r})$

$\boxed{m{\vec{a}}_{\mathrm{rot}}={\vec{F}}_{\mathrm{real}}\underset{{\vec{F}}_{\mathrm{Coriolis}}}{\underbrace{-2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}}}\underset{{\vec{F}}_{\mathrm{centrifugal}}}{\underbrace{-m\vec{\omega }\times (\vec{\omega }\times \vec{r})}}}$

Term by term:

• ${\vec{F}}_{\mathrm{real}}$ — genuine pushes/pulls (ropes, gravity, seat), dono observers ke liye same.
• $-2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$ — Coriolis pseudo-force: motion ${\vec{v}}_{\mathrm{rot}}$ chahiye, uske $\perp$ baith jaati hai, direction reverse karo toh flip ho jaati hai.
• $-m\vec{\omega }\times (\vec{\omega }\times \vec{r})$ — centrifugal pseudo-force: sirf kahan ho uspe depend karta hai, hamesha present.

PICTURE. Figure baithi observer ka free-body dikhata hai: real force inward, plus do invented arrows, balance karke woh acceleration dete hain jo woh actually measure karti hai.

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Step 7 — Centrifugal kis taraf point karta hai? (double cross product)

KYA. $-\vec{\omega }\times (\vec{\omega }\times \vec{r})$ ko ek plain direction mein decode karo.

KYUN. Ek cross product ke andar ek cross product scary lagta hai. Hum ise geometrically unpack karte hain taaki tum bina algebra ke "axis se bahar" dekh sako.

• Inner: $\vec{\omega }\times \vec{r}$ circle ke along (tangent) point karta hai, length $\omega r_{\perp }$, jahan $r_{\perp }$ axis se door hai.
• Outer: $\vec{\omega }$ ko us tangent arrow se cross karna use aur 90° rotate karta hai — woh axis ki taraf inward land hota hai, length ${\omega }^2{r}_{\perp }$.
• Minus sign use flip karta hai: outward, axis se door.

PICTURE. Do nested right-angle turns: $\vec{r}$ → (cross $\vec{\omega }$) → tangent → (cross $\vec{\omega }$) → inward → (minus) → outward. Har 90° turn draw kiya gaya hai.

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Step 8 — Edge & degenerate cases (kabhi surprise mat khaana)

KYA. Har special input sweep karo taaki koi scenario tumhe surprise na kare.

Case	${\vec{v}}_{\mathrm{rot}}$	Position	Coriolis	Centrifugal
Disc pe still khade hain	$\vec{0}$	off-axis	$\vec{0}$ (velocity chahiye)	outward, $m{\omega }^2{r}_{\perp }$
Axis pe baithe hain	$\vec{0}$	$r_{\perp }=0$	$\vec{0}$	$\vec{0}$
$\vec{\omega }$ ke parallel move karna*	$\parallel \vec{\omega }$	koi bhi	$\vec{0}$ ($\sin \varphi =0$)	outward as usual
Radially move karna (andar/bahar)	$\perp \vec{\omega }$	koi bhi	full, sideways	outward
Spin nahi ($\omega =0$)	koi bhi	koi bhi	$\vec{0}$	$\vec{0}$ (frame inertial hai!)

KYUN har ek vanish hota hai.

• Coriolis ko cross product nonzero chahiye: $|\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}|=\omega v\sin \varphi$. Agar ${\vec{v}}_{\mathrm{rot}}=0$ hai, ya ${\vec{v}}_{\mathrm{rot}}\parallel \vec{\omega }$ (toh $\varphi =0$), yeh zero hai.
• Centrifugal ko $r_{\perp }\ne 0$ chahiye: axis pe kuch nahi hai jisse flung ho sako.
• Agar $\omega =0$ toh frame spin nahi karta — yeh ek inertial frame hai — aur dono fictitious forces gayab ho jaati hain, jaise hona chahiye.

PICTURE. Chaar mini-panels: still rider (sirf outward arrow), axis rider (koi arrows nahi), radial thrower (sideways Coriolis), aur axis-parallel mover (no Coriolis). Har ek mein label hai ki kaun si force survive karti hai.

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Ek-picture summary

Yeh final panel poori ladder compress karta hai:

$\vec{r}\underset{\text{transport}}{\overset{d/dt}{\to }}{\vec{v}}_{\mathrm{in}}={\vec{v}}_{\mathrm{rot}}+\vec{\omega }\times \vec{r}\underset{\text{transport}}{\overset{d/dt}{\to }}{\vec{a}}_{\mathrm{in}}={\vec{a}}_{\mathrm{rot}}+2\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}+\vec{\omega }\times (\vec{\omega }\times \vec{r})$

$\Rightarrow m{\vec{a}}_{\mathrm{rot}}={\vec{F}}_{\mathrm{real}}-\underset{\text{Coriolis: motion ko turn karta hai}}{\underbrace{2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}}}-\underset{\text{centrifugal: bahar flings karta hai}}{\underbrace{m\vec{\omega }\times (\vec{\omega }\times \vec{r})}}$

Yeh numbers D3 Worked Examples mein le jao, phir D4 Exercises mein khud ko test karo. Accumulation effect ko action mein dekho Coriolis Effect in Weather aur Foucault Pendulum mein.

Recall Feynman retelling — plain words mein poori walkthrough

Do dost ek spinning merry-go-round dekhte hain. Ek upar float karta hai (spinning nahi), ek uspe sawari karta hai. Woh floor pe do arrows paint karte hain rulers ki tarah, plus ek teesra seedha floor se upar point karta hua. Rider ke liye rulers still baithe hain; floater ke liye woh sweep karte hain — aur us sweep ki speed hai "spin arrow crossed with ruler," ek perpendicular nudge (Step 2). Kyunki sab kuch un rulers se bana hai, floater hamesha rider ka change dekhta hai plus rulers ka sweep — yahi transport theorem hai (Step 3). Ise ek baar position pe apply karo → velocities differ karti hain "floor tumhe drag karta hai" se (Step 4). Ise phir velocity pe apply karo → accelerations differ karti hain do leftover terms se, aur unme se ek do baar show up hota hai, famous factor of 2 deta hua (Step 5). Newton sirf floater ki acceleration pe trust karta hai; jab rider insist karti hai force = mass × uski acceleration use karne par, woh do leftovers force side pe sneak ho jaate hain make-believe forces ki tarah (Step 6): centrifugal, jo hamesha tumhe axis se seedha bahar flings karta hai (Step 7), aur Coriolis, jo sirf tab bolta hai jab tum move karo aur hamesha tumhe sideways dhakela deta hai. Still khade raho → no Coriolis. Axis pe baitho → bilkul koi forces nahi. Spin band karo → poora ghost show gayab ho jaata hai aur Newton plain ho jaata hai (Step 8). Agar spin rate change kare, toh ek aur ghost — Euler force — party mein join karta hai, lekin humne spin steady rakha toh woh ghar pe hi raha.

Recall Quick self-test

Coriolis mein 2 ka factor aata hai ::: do identical $\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$ terms se jo doosri baar differentiate karne par appear hote hain. Centrifugal magnitude equal hoti hai ::: $m{\omega }^2{r}_{\perp }$, axis se outward point karte hue. Coriolis zero hoti hai jab ::: tum rotating frame mein rest par ho, ya $\vec{\omega }$ ke parallel move karo, ya $\omega =0$ ho. Centrifugal zero hoti hai jab ::: tum rotation axis pe baithe ho ($r_{\perp }=0$) ya $\omega =0$ ho. Extra pseudo-force jo sirf tab present hoti hai jab spin rate change ho, woh hai ::: Euler force $-m\dot{\vec{\omega }}\times \vec{r}$.