Visual walkthrough — Constraints — holonomic vs non-holonomic, rheonomic vs scleronomic

2.1.1 · D2 · Physics › Analytical Mechanics › Constraints — holonomic vs non-holonomic, rheonomic vs scler

Step 1 — "Position" ka matlab kya hai ek coin ke liye table par

KYA HAI: Hum coin ki state ko $(x,y,\varphi ,\theta )$ se naam dete hain.

KYUN: Isse pehle ki hum poochhen "kya change ho sakta hai?", pehle yeh kehna zaroori hai ki "change hone ke liye hai kya." Chaar independent knobs = chaar dimensions ka ek configuration space.

PICTURE: Amber dot dekho — contact point $(x,y)$ par. Cyan arrow heading $\varphi$ hai. Rim par chhota sa wedge $\theta$ track karta hai, yaani roll.

Step 2 — Ek hi rule: no slipping

Radius $R$ ke coin ke liye, rim ka angle $d\theta$ ek arc hai jisme length $Rd\theta$ hai. Toh contact point $Rd\theta$ distance aage badhta hai jis direction mein coin point kar raha ho.

KYA HAI: Hum "no slipping" ko ek precise length mein convert karte hain: distance moved $=Rd\theta$.

KYUN: Humein physics ("no skid") ko numbers ke roop mein likhna hai jo tiny changes relate karte hain, kyunki tiny changes hi woh cheez hain jo constraints control karti hain.

PICTURE: Rim wedge $d\theta$ table par amber segment ke roop mein unroll hota hai jisme length $Rd\theta$ hai.

Step 3 — Us motion ko East aur North mein split karna

KYA HAI: Hum step $ds=Rd\theta$ ko $dx$ aur $dy$ mein resolve karte hain.

KYUN: Humne $\cos ,\sin$ choose kiya (aur $\tan$ nahi) kyunki hum ek known-length step ke actual East aur North lengths chahte hain ek known angle par — aur cosine/sine precisely wahi ratios hain jo woh lengths return karte hain.

PICTURE: Amber hypotenuse step hai; cyan legs uske East aur North shares hain.

Step 4 — Roll ko erase karke pure direction rule dekhna

Dono equations ko divide karo (ya cross-multiply karo):

KYA HAI: Humne $d\theta$ hataya aur $dx,dy,\varphi$ ke beech ek single relation paya.

KYUN: Yeh constraint ki sharp form hai: contact point sirf heading ke along slide kar sakta hai, kabhi sideways nahi. Sideways motion $dx\sin \varphi -dy\cos \varphi$ forbidden hai.

PICTURE: Green arrow = allowed (along $\varphi$). Red arrow = woh forbidden sideways slide jise equation kill karta hai.

Step 5 — Integrability test: kya yeh position rule ban sakta hai?

Agar $Adx+Bdy=0$ kisi $f$ ke liye $df$ hota, toh $A=\partial f/\partial x$ aur $B=\partial f/\partial y$ hota, aur kyunki mixed partials agree karte hain, humein exactness condition chahiye hoti:

KYA HAI: Hum test karte hain ki $A=\sin \varphi ,B=-\cos \varphi$ exactness pass karte hain ya nahi — lekin note karo ki $A,B$ $\varphi$ par depend karte hain, jo ek teesra variable hai, isliye humein $d\varphi$ direction bhi allow karni hogi. Teen variables mein honest test Frobenius condition hai $\mathbf{a}\cdot (\nabla \times \mathbf{a})=0$, jahan $\mathbf{a}=(\sin \varphi ,-\cos \varphi ,0)$ hai.

KYUN: Humein ek yes/no machine chahiye. Exactness (2 variables) ya Frobenius (3 variables) woh machine hai: yeh detect karta hai ki koi integrating factor kabhi exist kar sakta hai ya nahi.

PICTURE: Left panel — ek exact field jiske arrows sab ek landscape $f$ ke "downhill" hain (integrable). Right panel — hamaara rolling field, jiske arrows swirl karte hain jab $\varphi$ turn karta hai, isliye koi single landscape nahi hai jiska slope woh arrows hon.

$\mathbf{a}=(\sin \varphi ,-\cos \varphi ,0)$ ka curl compute karo (treating $x,y,\varphi$ ko teen axes ki tarah):

Phir

Result: Test $-1$ return karta hai, kabhi zero nahi. Koi integrating factor exist nahi karta. Rolling rule kisi bhi $f(x,y,\varphi )$ ka differential nahi hai. → Non-holonomic.

Step 6 — Wandering se proof: phir bhi kisi bhi configuration tak pahuncho

KYA HAI: Hum dikhate hain ki reachable configurations saari hain, velocity rule ke bawajood.

KYUN: Yahi "non-integrable" ka physical meaning hai: agar ek position equation $f=0$ exist karti, toh tum ek surface par hamesha ke liye stuck rehte. Tum stuck nahi ho ⇒ aisi koi surface nahi ⇒ confirms non-holonomic, bahar se.

PICTURE: Ek "parallel-park" loop: aage roll karo, steer karo, roll karo, wapas steer karo — coin apni starting position se sideways shift ho jaata hai, ek aisa move jo instantaneous rule "forbid" karta tha. Yeh net sideways shift non-holonomy ki signature hai.

Step 7 — Time axis: kya yeh scleronomic hai?

KYA HAI: Hum confirm karte hain ki koi explicit time nahi hai.

KYUN: Classification ke dono axes independent hain (parent ke second mistake dekho). Axis 1 par non-holonomic settle karne ke baad, axis 2 ek alag yes/no hai.

PICTURE: Ek frozen table — constraint surface time ke saath breathe nahi karta. Parent ke spinning wire ($y=x\tan \omega t$) se compare karo, jisme $t$ hoga.

Verdict: rolling coin non-holonomic aur scleronomic hai.

Ek-picture summary

Sab kuch ek page par: coin heading $\varphi$ ke along ek step $Rd\theta$ roll karta hai (top-left); woh step $dx=R\cos \varphi d\theta$, $dy=R\sin \varphi d\theta$ mein split hota hai (top-right); $d\theta$ eliminate karne par sideways motion forbid hoti hai, $\sin \varphi dx-\cos \varphi dy=0$ (bottom-left); swirl test $\mathbf{a}\cdot (\nabla \times \mathbf{a})=-1\ne 0$ deta hai, isliye koi $f$ exist nahi karta, lekin parallel-parking kisi bhi $(x,y,\varphi )$ tak pahuncha deti hai (bottom-right).