Visual walkthrough — Circular motion — centripetal acceleration derivation

1.2.15 · D2 · Physics › Newton's Laws & Dynamics › Circular motion — centripetal acceleration derivation

Yeh page parent derivation ka visual companion hai. Agar koi word naya lage, toh hum usse wahi se zero se build karte hain jahan woh aata hai.

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Step 1 — "Circle mein move karna" ka matlab kya hota hai

WHAT. Ek akela dot imagine karo jo ek circle ke edge pe ghoom raha hai. Dot kabhi slow ya fast nahi hota — uski speed (har second kitne metres cover karta hai) fixed hai. Hum us speed ko $v$ kehte hain (units: metres per second, $\text{m/s}$). Circle ka ek fixed radius $r$ hai — centre se edge tak ki doori (units: metres, $\text{m}$).

WHY. Hume "constant speed circular motion" ko precisely define karna hoga, kyunki is topic ki poori surprise yeh hai ki constant speed ka matlab bhi accelerate karna hai. Yeh dekhne ke liye pehle hume do ideas alag karni hogi jo rozaana ki "speed" word milaa deti hai:

• Speed = velocity arrow ki length (ek plain number).
• Velocity = ek arrow: ismein length AUR direction dono hote hain.

PICTURE. Niche, dot circle pe baitha hai. Dot se nikalta seedha arrow velocity $\vec{v}$ hai. $v$ ke upar chhota half-arrow, likh a $\vec{v}$, humara reminder hai ki yeh ek arrow-quantity hai, sirf number nahi. Note karo arrow tangent hai — yeh circle ko sirf graze karta hai, pointing the way the dot abhi jaane wala hai.

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Step 2 — Do snapshots, thodi si time ke baad

WHAT. Film ko do baar freeze karo. Pehle freeze mein dot point A pe hai; bahut thodi time baad — hum us gap ko $\Delta t$ kehte hain — woh point B pe hai. Symbol $\Delta$ (Greek "delta") ka matlab hai "mein ek chhota sa change." Toh $\Delta t$ = "elapsed time ki ek chhoti si maatra."

WHY. Acceleration time ke upar velocity ka change hai. Change measure karne ke liye humein ek before aur ek after chahiye. Do snapshots hume exactly woh dete hain: A pe velocity vs B pe velocity.

PICTURE. Centre se hum do spokes draw karte hain: ${\vec{r}}_A$ point A tak aur ${\vec{r}}_B$ point B tak. Dono spokes ki length $r$ same hai (yeh same circle hai). Unke beech ka wedge chhota angle $\Delta \theta$ ("delta theta") hai — woh turn ka slice jo dot ne $\Delta t$ ke dauran sweep kiya.

Term-by-term, A se B tak arc: $\underset{\text{curved distance travelled}}{\underbrace{\text{arc AB}}}=\underset{\text{radius}}{\underbrace{r}}\underset{\text{angle swept, in radians}}{\underbrace{\Delta \theta }}$

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Step 3 — Ek tiny wedge ke liye, chord ≈ arc

WHAT. Dot actually curved arc pe travel karta hai, lekin jab wedge patla ho toh straight line AB (chord) almost same length hoti hai.

WHY. Hum straight arrows compare karna chahte hain (velocities straight arrows hoti hain), isliye curve ko ek straight segment se replace karna helpful hai. Yeh sirf limit mein legal hai jab angle tiny ho — aur woh limit lena exactly wahi hai jo acceleration demand karta hai (ek instantaneous rate).

PICTURE. Dekho chord (deep-teal) arc (burnt-orange) ke kitna pass aati hai jaise jaise wedge narrow hota jaata hai. Woh shrinking gap poora trick hai.

$\underset{\text{straight chord length}}{\underbrace{|\text{AB}|}}\approx \underset{\text{arc length}}{\underbrace{r\Delta \theta }}(\text{exact as }\Delta \theta \to 0)$

$\approx$ ("approximately equals") ek true $=$ ban jaata hai jab hum $\Delta \theta$ ko kuch nahi hone dete.

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Step 4 — Velocity arrows SAME wedge banate hain

WHAT. Do velocity arrows lo, ${\vec{v}}_A$ (A pe velocity) aur ${\vec{v}}_B$ (B pe velocity). Dono ki length $v$ hai. Unhe slide karo taaki unki tails touch karein. Unke beech ka angle bhi $\Delta \theta$ hai — ekdam wahi wedge jo spokes ke beech tha.

WHY — key insight. Har velocity tangent hai, isliye woh apne spoke ke saath right angle ($90°$) pe baithti hai: ${\vec{v}}_A\perp {\vec{r}}_A$ aur ${\vec{v}}_B\perp {\vec{r}}_B$. Agar tum do arrows ko same $90°$ se rotate karo, toh unke beech ka angle unchanged rehta hai. Toh velocity pair spoke pair hi hai ek quarter turn ghuma ke — same enclosed angle $\Delta \theta$.

PICTURE. Left: spoke wedge. Right: velocity wedge, visibly same opening angle, bas rotated. Isliye woh do triangles jo hum compare karne wale hain woh similar hain (same shape, different size).

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Step 5 — Similar triangles humein $|\Delta \vec{v}|$ dete hain

WHAT. Dono triangles ko side by side rakho. Dono mein, ek lambi side aur ek chhoti "opening" side:

triangle	do lambi (equal) sides	chhoti opening side
position	$r$ aur $r$	chord $
velocity	$v$ aur $v$	$

Yahan $\Delta \vec{v}={\vec{v}}_B-{\vec{v}}_A$ velocity mein change hai — woh arrow jo ${\vec{v}}_A$ mein add karo toh ${\vec{v}}_B$ milti hai. Iski length $|\Delta \vec{v}|$ hai.

WHY. Similar triangles ka matlab hai dono mein short-side ÷ long-side identical hai. Yeh ek equation poori derivation ka pivot hai — yeh positions ke baare mein ek fact ko velocities ke baare mein fact mein convert karti hai.

PICTURE. Plum arrow $\Delta \vec{v}$ hai, woh missing third side jo velocity triangle ko close karti hai.

$\underset{\begin{array}{c}\text{short/long in}\\ \text{velocity triangle}\end{array}}{\underbrace{\frac{|\Delta \vec{v}|}{v}}}=\underset{\begin{array}{c}\text{short/long in}\\ \text{position triangle}\end{array}}{\underbrace{\frac{|\text{AB}|}{r}}}=\frac{r\Delta \theta }{r}=\Delta \theta$

Dono sides ko $v$ se multiply karo: $\boxed{|\Delta \vec{v}|=\underset{\text{speed}}{\underbrace{v}}\underset{\text{angle swept}}{\underbrace{\Delta \theta }}}$

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Step 6 — Time se divide karo, phir shrink karo: the limit

WHAT. Acceleration magnitude hai "velocity arrow kitni tezi se change hoti hai," yaani $|\Delta \vec{v}|$ per unit time. $\Delta t$ se divide karo: $\frac{|\Delta \vec{v}|}{\Delta t}=\frac{v\Delta \theta }{\Delta t}=v\frac{\Delta \theta }{\Delta t}$

WHY the limit. Ratio $\frac{\Delta \theta }{\Delta t}$ hai "angle swept per second" — lekin sirf $\Delta t$ pe averaged. Instantaneous acceleration paane ke liye hum $\Delta t\to 0$ squeeze karte hain. Woh squeezing limit hai, likha $\lim$. Jaise gap vanish hota hai, $\frac{\Delta \theta }{\Delta t}$ turning ka exact rate ban jaata hai, jise angular velocity $\omega$ kehte hain (dekho Angular velocity and period).

PICTURE. Bar chart dikhata hai $\frac{|\Delta \vec{v}|}{\Delta t}$ ek single value pe settle hota jaata hai jaise $\Delta t$ shrink hota hai — wobbling band ho jaata hai aur lock in ho jaata hai.

$a_c={\lim }_{\Delta t\to 0}\frac{|\Delta \vec{v}|}{\Delta t}=v\underset{=\omega }{\underbrace{\underset{\Delta t\to 0}{\lim }\frac{\Delta \theta }{\Delta t}}}=v\omega$

Ab speed aur turning-rate ke beech bridge use karo, $v=\omega r$ (radius $r$ pe ek dot $\omega$ radians/s se turn karta hai toh $v=\omega r$ metres/s sweep karta hai): $\boxed{a_c=v\omega =\frac{v^2}{r}={\omega }^{2}r}$

Har swap sirf $v=\omega r$ use karta hai: $\omega =v/r$ daalo toh $v^2/r$ milta hai, ya $v=\omega r$ daalo toh ${\omega }^{2}r$ milta hai.

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Step 7 — Yeh point kis direction mein karta hai? (direction)

WHAT. Jaise $\Delta t\to 0$, arrow $\Delta \vec{v}$ tab tak turn karta hai jab tak woh $\vec{v}$ ke perpendicular aur centre ki taraf point karne wala na ho jaaye.

WHY. Velocity triangle mein do equal sides dono length $v$ ki hain. Jab opening angle $\Delta \theta$ zero ho jaata hai, ek isosceles triangle ki base uski equal sides ke perpendicular ho jaati hai. Geometrically woh base ($\Delta \vec{v}$) seedha andar aim karne lagti hai — isliye centre-seeking, "centripetal." Ek inward-bending velocity exactly woh hai jise Newton's second law phir ek real inward force se connect karta hai.

PICTURE. Jaise wedge close hota hai, dekho $\Delta \vec{v}$ swing karta hai jab tak woh directly centre ki taraf point nahi karta.

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Step 8 — Edge & degenerate cases (koi gap mat chhodna)

WHAT / WHY / PICTURE — har limit check kiya.

• $v=0$ (dot move nahi kar raha). Toh $a_c=0^2/r=0$. Ek parked dot ko koi inward pull nahi chahiye. ✔
• $r\to \infty$ (circle ek line mein seedha ho jaata hai). $a_c=v^2/r\to 0$. Ek giant circle locally ek straight road hai — koi bending nahi, koi centripetal acceleration nahi. Isliye ek gentle bend easy lagta hai. ✔
• $r\to 0$ (spot pe spin karo). $a_c=v^2/r\to \infty$ any $v>0$ ke liye: ek impossibly tight turn infinite acceleration demand karta hai. Isliye real turns perfectly sharp nahi ho sakte. ✔
• Quadrant check (circle ke chaaron taraf). Calculus form $\vec{a}=-{\omega }^2\vec{r}$ har point pe kaam karta hai: top, bottom, left, right. Minus sign guarantee karta hai ki $\vec{a}$ hamesha outward spoke ke opposite point kare — yaani inward — sab four quadrants mein, koi special cases nahi. ✔

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Ek-picture summary

Sab ek saath: position wedge → $90°$ rotate karo → velocity wedge → similar-triangle ratio → time se divide karo → limit → $a_c=v^2/r$, pointing inward.

Recall Feynman retelling — poori film simple words mein

Ek dot ek circle ke around steady pace se dauda raha hai. Main do photos leta hoon ek heartbeat apart: A pe aur B pe. A aur B ke do spokes ek patla angle ka slice kholte hain. Kyunki dot ki velocity arrow hamesha apne spoke ke right angle pe baithti hai, do velocity arrows exactly same patla slice kholte hain — bas ek quarter-turn ghuma ke. Toh velocities se bana chhota triangle spokes se bane triangle ka shrunken twin hai. Twins side-ratios share karte hain, aur woh ratio mujhe velocity-change arrow ki length batata hai: $|\Delta \vec{v}|=v\Delta \theta$. Heartbeat se divide karo aur heartbeat ko kuch nahi tak shrink karo, aur mujhe milta hai $a_c=v\omega =v^2/r$. Aakhir mein, jaise slice close hota hai, woh velocity-change arrow ghoom ke seedha centre ki taraf point karne lagta hai. Toh: constant speed, phir bhi hamesha accelerating — inward, forever, at $v^2/r$.

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Recall

Velocity vectors usi angle $\Delta \theta$ ko kyun enclose karte hain jitna radius vectors? ::: Har velocity apne radius ke perpendicular hai; dono radii ko $90°$ se rotate karke velocities banane se unke beech ka angle preserve hota hai. Limit $\Delta t\to 0$ mein, $\Delta \theta /\Delta t$ kya ban jaata hai? ::: Angular velocity $\omega$, radians per second mein instantaneous turning rate. $v$ fixed rakh ke $r\to \infty$ hone par $a_c$ ka kya hoga, aur physically kyun? ::: $a_c\to 0$; ek huge circle locally straight hai, toh velocity barely bend hoti hai. $\vec{a}=-{\omega }^2\vec{r}$ mein minus sign kya guarantee karta hai? ::: Acceleration outward radius ke opposite point karta hai — yaani center ki taraf — har quadrant mein.