Visual walkthrough — Relation to linear quantities - v = rω, a_t = rα, a_c = rω²

1.5.3 · D2 · Physics › Rotational Mechanics › Relation to linear quantities - v = rω, a_t = rα, a_c = rω²

Hum do bhaashaaon ke beech translate kar rahe hain: angular wali (kitna ghuma) aur linear wali (kitni door aur kitni tez circle ke saath chale). Poora bridge ek hi shabd mein hai: radius $r$.

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Step 1 — Arc draw karo: "$r$", "$\theta$", "$s$" actually hain kya

KYA. Ek ghoomti hui disk ke center par ek dot rakho — use axis kaho. Disk par ek point $P$ chuno jo axis se $r$ door hai. Jaise-jaise disk ghoomti hai, $P$ ek circle ka ek tukda sweep karta hai.

YEH TEEN KYU. Is page par sab kuch inhi se bana hai. Speed aayegi "kitni tez $s$ badhta hai?" se; angular quantities aayengi "kitni tez $\theta$ badhta hai?" se. Isliye pehle tasveer pakka karni hogi.

TASVEER. Neeche wala blue wedge: radius $r$ (blue), swept angle $\theta$ (orange), aur curved arc $s$ (green) rim ke saath chalta hua.

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Step 2 — Radians kyun: equation $s=r\theta$

KYA. Hum claim karte hain ki arc length is rule ko follow karta hai: $s=r\theta$ Term-by-term padho: $s$ (curved distance) barabar hai $r$ (tum kitne door ho) guna $\theta$ (tune kitna angle sweep kiya).

RADIANS kyun, degrees nahi. Ek radian ko define kiya gaya hai us angle ke roop mein jiske liye arc, radius ke barabar hoti hai. Theek $1$ radian ghoomne par point exactly ek $r$ jitni curved distance chalti hai. Toh "radians mein angle" ka matlab literally hai "maine kitni radius-lengths ki arc trace ki." Yahi poora trick hai — yeh $s=r\theta$ ko bina kisi conversion factor ke clean banata hai. Degrees mein tumhe $s=r\theta \cdot \frac{\pi }{180}$ likhna padta, ek bekar extra kaam jo hum nahi karna chahte. (Aur detail mein Radian measure and arc length mein.)

TASVEER. Usi spoke par do dots: ek andar wala ($r$ chhota) aur ek bahar wala ($r$ bada). Same $\theta$, lekin bahar wala arc clearly zyada lamba hai — kyunki $s$, $r$ ke saath scale karta hai. Yahi reason hai ki rim zyada tez chalti hai; hum ise abhi formula banana wale hain.

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Step 3 — Arc se speed tak: $v=r\omega$

KYA. Ab hum poochhte hain: point apne circle ke saath kitni tez move karta hai? Matlab: $s$ kitni tez badh raha hai? "Kisi quantity ka rate of change" exactly wahi hai jo derivative measure karta hai — isliye woh tool yahan aata hai. $\frac{ds}{dt}$ likhne ka matlab hai "time $t$ ke chhote se tick mein $s$ kitna badla."

DERIVATIVE kyun, sirf $s/t$ kyun nahi. Agar spin bilkul steady hoti, toh $s/t$ kaam karta. Lekin hum ek aisa rule chahte hain jo har instant kaam kare, chahe spin up ho raha ho. Derivative $s/t$ ka "is exact moment par" wala version hai. Ise $s=r\theta$ par apply karo: $\frac{ds}{dt}=\frac{d}{dt}\left(r\theta \right)=r\frac{d\theta }{dt}$ Humne $r$ aage nikala kyunki $r$ constant hai (rigid disk). Agar $r$ badal sakta toh ek extra piece aata — Step 6 dekho.

TASVEER. Velocity arrow $\vec{v}$ circle ke tangent par drawn (spoke $r$ se right angle par). Double radius par ek twin arrow exactly double lamba hai — same $\omega$, double $r$, double $v$.

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Step 4 — Spin ko tez karo: $a_t=r\alpha$

KYA. Ab disk ko sirf spin nahi, balki zyada se zyada tez spin karne do. Tab $\omega$ khud time ke saath badlta hai. Jis rate par $\omega$ badlta hai woh hai angular acceleration $\alpha$ (alpha): $\alpha =\frac{d\omega }{dt}({\text{rad/s}}^2).$

DOBARA DIFFERENTIATE kyun. Speed $v$ ab isliye badlti hai kyunki $\omega$ badal raha hai. "Speed ka rate of change" hai tangential acceleration $a_t$ — phir se ek derivative, pehle wale jaisi hi wajah se: hum instantaneous rule chahte hain. $v=r\omega$ ko differentiate karo ($r$ abhi bhi constant hai): $\frac{dv}{dt}=r\frac{d\omega }{dt}\Rightarrow a_t=r\alpha$

Edge case yahan: agar disk steadily spin kare, $\omega$ constant hai, toh $\alpha =0$ aur isliye $a_t=0$. Bilkul bhi tangential acceleration nahi — yeh exactly uniform circular motion hai.

TASVEER. Orange ${\vec{a}}_t$ arrow $\vec{v}$ ke same direction mein point karta hua (disk speed up ho rahi hai), rim ke tangent par baitha hua.

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Step 5 — Woh sneaky wala: constant speed par bhi circle accelerate kyun karta hai

KYA. Yahan surprise hai. Chahe point bilkul constant speed par move kare, tab bhi woh accelerate kar raha hota hai — kyunki acceleration velocity as a vector ki parwah karta hai, aur velocity ki direction lagatar ghoomti rehti hai.

VECTORS kyun, sirf speed nahi. Speed ek single number hai; velocity $\vec{v}$ ek aisa arrow hai jisme length aur direction dono hote hain. Acceleration ka matlab hai "velocity arrow kaise badlti hai." Circle par arrow ki length fixed reh sakti hai jabki uski heading ghoomti rehti hai — woh change real hai aur kisi acceleration se honi chahiye. Isliye ab hum position vector ko differentiate karte hain, sirf $s$ ko nahi.

TASVEER. Circle ke chaar points par $\vec{v}$ ke chaar snapshots — arrow same length rakhta hai lekin chaar alag directions mein point karta hai. Woh ghoomti hui heading wahi cheez hai jo hum abhi differentiate karne wale hain.

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Step 6 — Vector ko do baar differentiate karo: $a_c=r{\omega }^2$

KYA. Position ko ek baar differentiate karo velocity ke liye, do baar acceleration ke liye. (Calculus facts yaad karo: $\cos \theta$ ka slope $-\sin \theta$ hai, $\sin \theta$ ka slope $\cos \theta$ hai, aur $\theta =\omega t$ har baar chain rule se ek factor $\omega$ contribute karta hai.)

Velocity (pehla derivative): $\vec{v}=r\omega (-\sin \theta ,\cos \theta ).$ Iski length $r\omega$ hai — Step 3 se match karta hai — aur yeh $\vec{r}$ se right angle par point karta hai (tangent). Achha, consistent hai.

Acceleration (doosra derivative): $\vec{a}=r{\omega }^2(-\cos \theta ,-\sin \theta )=-{\omega }^2\vec{r}.$

• Bracket $(-\cos \theta ,-\sin \theta )$ exactly $\vec{r}$ ulta hai — yeh $P$ se center ki taraf point karta hai.
• Minus sign yahi headline hai: acceleration andar ki taraf aim karta hai. Isliye ise centripetal ("center-seeking") kehte hain.

$r$ ko pehle kyun nikala — ab fayda milta hai: Steps 3–4 mein humne $r$ constant assume kiya tha. Step 6 dikhata hai ki "constant $r$" se kya milta hai: ek pure inward acceleration bina kisi messy in/out drift term ke. Agar $r$ badlta (ek bead ghoomte rod par slide kare), toh extra terms aate — woh advanced motion problems mein hai, is page se aage.

TASVEER. Red ${\vec{a}}_c$ arrow dead center ki taraf point karta hua, tangential arrows ke saath drawn taaki tum dekh sako ki woh right angles par hain.

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Step 7 — Do accelerations ko saath rakhna

KYA. Ek general point ke paas dono accelerations ek saath hoti hain: $a_t$ (tangent, speed badlti hai) aur $a_c$ (andar, direction badlti hai). Kyunki woh 90° par point karti hain, total magnitude ke liye Pythagoras use karte hain: $a=\sqrt{a_t^2+a_c^2}.$

PYTHAGORAS kyun. Do perpendicular arrows ek right triangle ki legs ki tarah add hoti hain; resultant hypotenuse hota hai. Koi cross-term nahi kyunki $\cos 90°=0$.

TASVEER. ${\vec{a}}_t$ (leg), ${\vec{a}}_c$ (leg), aur total $\vec{a}$ (hypotenuse) ka right triangle, point par drawn.

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Step 8 — Har case check kiya

Recall Saare corner cases walk through karo

• Axis par point ($r=0$): $v=0$, $a_t=0$, $a_c=0$. Center move nahi karta — sahi hai.
• Steady spin ($\alpha =0$): $a_t=0$, lekin $a_c=r{\omega }^2\ne 0$. Constant speed, phir bhi ghoom raha hai. Uniform circular motion.
• Momentarily at rest lekin wind up ho raha hai ($\omega =0,\alpha \ne 0$): $a_c=0$ (abhi redirect karne ke liye koi speed nahi) lekin $a_t=r\alpha \ne 0$. Us instant par saari acceleration tangential hai.
• Slow down ho raha hai ($\alpha <0$): $a_t$ flip hoke $\vec{v}$ ke against point karta hai; $a_c$ abhi bhi andar point karta hai. Formulas mein sign already hai.
• $\omega$ double karo, $r$ aur $\alpha$ fix rakho: $v$ double hota hai, $a_t$ unchanged (usme kabhi $\omega$ tha hi nahi!), $a_c$ chaar guna hota hai. Parent note wala $a_t$-trap.

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Ek tasveer mein summary

Is page par sab kuch ek arc se bada. Neeche wali figure ise compress karti hai: arc $\to$ speed $\to$ do accelerations, har ek ke saath uska formula aur woh tool jo ise banata hai.

Recall Feynman retelling — poora walkthrough seedhe shabdon mein

Ek pizza ka slice draw karo: tip axis hai, crust par ek point tumhara traveller hai. Radius $r$ hai ki traveller kitna door baitha hai; angle $\theta$ (radians mein, is liye choose kiya ki ek radian ghoomne par exactly ek radius crust ke saath chalo) bata hai kitna ghuma; arc $s=r\theta$ curved distance hai jo chali.

Poochho "crust ke saath kitni tez?" — woh speed hai, $s$ ki growth rate, aur kyunki $r$ kabhi nahi badlta, yeh sirf $r$ guna angle ki growth rate hai, yaani $v=r\omega$. Door hona matlab same ghoomne ke liye lambi arc, isliye rim tez hai.

Ab pizza ko zyada se zyada tez ghoomao. Angle ki growth-rate khud badlti hai; use $\alpha$ kaho. Traveller ki speed rate $a_t=r\alpha$ se badlti hai — crust ke saath aage (ya peeche) ka dhakka.

Aakhir mein, sneaky part: bilkul steady spin par bhi, traveller poore time tip ki taraf khicha ja raha hota hai, kyunki jis taraf woh face kar raha hai woh lagatar ghoomti rehti hai. Uski position $(r\cos \theta ,r\sin \theta )$ ke roop mein track karo, do baar differentiate karo, aur size $a_c=r{\omega }^2$ ka ek andar wala pull nikalti hai. Spin double karo aur yeh pull chaar guna strong ho jaata hai — exactly isliye tez ghoomne wali cheezein khud ko bikhar ne ki koshish karti hain. Do accelerations right angle par hain, isliye total hypotenuse $\sqrt{a_t^2+a_c^2}$ hai. Ek arc, teen formulas, ek dictionary word: $r$.