Banking of roads — derivation: Visual walkthrough

1.2.17 · D2 · Physics › Newton's Laws & Dynamics › Banking of roads — derivation

Hum ek relationship dhundh rahe hain: ek curved road ko kitna tilt karna chahiye taaki speed $v$ se radius $r$ ke circle mein jaata hua ek car apne path par reh sake — bilkul bhi grip na ho tab bhi?

Step 1 — Problem: ek turn ke liye ek sideways pull chahiye

KYA. Ek car ek flat circular track par drive kar rahi hai. Uski velocity arrow dekho: direction tab bhi badalta rehta hai jab speed same rehti hai.

KYU. Jo bhi cheez circle mein move karti hai wo constantly turn ho rahi hoti hai. Turning matlab velocity vector rotate karta hai, aur ek rotating velocity ek acceleration hai — jo seedha circle ke centre ki taraf point karta hai. Yahi centripetal ("centre-seeking") acceleration hai. Ise cause karne ke liye koi real force car ko inward push karni chahiye. Yeh kahan se aata hai, dekho Uniform Circular Motion.

PICTURE. Car (top view) apni velocity arrow ke saath circle ke tangent par, aur ek red arrow centre ki taraf point karta hua — wo force jo hume kisi na kisi tarah supply karni hai.

Step 2 — Flat ground par, sirf friction hi yeh kar sakta hai (aur yahi khatre ki baat hai)

KYA. Ek flat road par single horizontal force jo available hai wo tyre aur road ke beech friction hai. Toh friction akele $\frac{m v ^{2}}{r}$ ke barabar hona chahiye.

KYU. Weight seedha neeche point karta hai, road ka push (normal force) seedha upar — dono vertical hain. Dono mein koi sideways part nahi hai. Ek force jiske paas horizontal component hai wo Friction hai. Agar road geeli, barf wali, ya taili hai, toh friction kam ho jaata hai aur car curve se bahar slide ho jaati hai.

PICTURE. Flat ground par car ka side view: weight neeche, normal upar (dono vertical, cancel kar rahe hain), aur ek akela horizontal friction arrow inward point karne ki koshish kar raha hai.

Step 3 — Road ko tilt karo: normal force lean karna seekh jaati hai

KYA. Road ki outer edge ko angle $θ$ se utha lo. Road surface ab ek slope hai. Normal force — road ka push, jo hamesha surface ke perpendicular hota hai — ab seedha upar point nahi karta. Woh usi angle $θ$ se inward tilt ho jaata hai.

KYU perpendicular? Ek surface sirf push kar sakta hai, kabhi bhi ideal frictionless contact mein sideways pull nahi kar sakta — toh push surface ke right angles par hota hai. Surface ko $θ$ se tilt karo, aur perpendicular bhi $θ$ se tilt ho jaata hai. Ab normal force ne ek horizontal, inward-pointing component develop kar liya hai — bilkul wahi direction jo hume Step 1 mein chahiye thi.

PICTURE. Banked road ka cross-section. Normal force $N$ (blue) vertical se banking angle $θ$ (yellow) se lean karta hai, jabki weight $m g$ (red) ab bhi seedha neeche hang karta hai.

Step 4 — $N$ ko ek vertical part aur ek horizontal part mein split karo

KYA. Hum tilted arrow $N$ ko do arrows mein tod te hain jo right angles par hain: ek seedha upar, ek seedha sideways (inward).

KYU yeh do directions? Kyunki motion khud unhi ke along cleanly split hoti hai. Vertically car accelerate nahi karti — woh na road mein ghusati hai na upar uthti hai, toh vertical forces balance karne chahiye. Horizontally car accelerate karti hai centre ki taraf. "Up" aur "inward" ko apne axes choose karne se hum ek direction per ek honest equation likh sakte hain. (Same trick jaise Inclined plane mein.)

PICTURE. $N$ par bana ek right triangle: hypotenuse $N$ khud hai, vertical leg $N cos θ$ hai, horizontal leg $N sin θ$ hai.

Step 5 — Do balance equations likho (frictionless case)

KYA. Abhi ke liye perfectly slippery ice assume karo — friction $f = 0$ . Ek equation up–down aur ek equation in–out likho.

KYU. Bina friction ke, sirf $N$ aur gravity act karte hain. Yeh sabse clean possible result deta hai — woh "design" case jiske liye road banayi jaati hai.

Vertical — kuch bhi up ya down accelerate nahi karta: $\underbrace{N\cos\theta}_{\text{road ke push ka up-part}} \;=\; \underbrace{mg}_{\text{weight jo neeche pull karta hai}} \tag{1}$ $N$ ka upward slice gravity ko exactly cancel karna chahiye, warna car sink ya fly kar jaayegi.

Horizontal — yahi centripetal force hai: $\underbrace{N\sin\theta}_{\text{road ke push ka inward slice}} \;=\; \underbrace{\frac{mv^2}{r}}_{\text{inward force jo turn demand karta hai}} \tag{2}$ $N$ ka inward slice hi poori centripetal force hai — koi friction help nahi kar raha.

PICTURE. Do triangle legs apne kaam se colour-matched: vertical leg $m g$ se lad rahi hai, horizontal leg $\frac{m v ^{2}}{r}$ supply kar rahi hai.

Step 6 — Equations divide karo: magic cancellation

KYA. Equation (2) ko equation (1) se divide karo, left side over left, right side over right.

KYU divide karo? Do unknowns hamare equations ko cluttered karte hain: normal force $N$ aur mass $m$ . Dono aise cheez nahi hain jo hum road design karte waqt choose kar sakte hain. Divide karne se dono ek saath gayab ho jaate hain — $N$ left side par cancel hota hai, $m$ right side par cancel hota hai — sirf wahi quantities bachti hain jo ek road engineer actually control karta hai: angle, speed, radius, gravity.

$\frac{N s i n θ}{N c o s θ} = \frac{m v ^{2} / r}{m g}$

$N$ cancel ho jaate hain; $m$ cancel ho jaate hain. Aur $\frac{s i n θ}{c o s θ}$ by definition $tan θ$ hai — hamare triangle ka "steepness ratio", unit up ke saath kitna sideways.

Step 7 — Friction wapas aata hai: jab tum zyada fast jaate ho toh friction NEECHE point karta hai

KYA. Ab friction allow karo. Agar car design speed se zyada fast jaaye, toh woh up-and-outward slide karne ki koshish karti hai. Friction us tendency ko oppose karta hai, toh woh slope se neeche act karta hai, apna khud ka inward help add karta hua — ek higher top speed $v_{ma x}$ allow karta hua.

KYU slope se neeche? Friction hamesha us direction se ladta hai jis mein surface slip karna chahta hai. Zyada fast ⇒ car outward climb karna chahti hai ⇒ surface outward-up slip karna chahta ⇒ friction downward-inward grip karta hai. Yeh sabse bada value le sakta hai jo $f = μ N$ hai, jahan $μ$ = coefficient of Friction.

Vertical (down-slope friction ka ek downward slice bhi hai): $N\cos\theta = mg + \underbrace{\mu N\sin\theta}_{\text{friction ka down-part}} \tag{3}$

Horizontal ( $N$ aur friction dono ab inward push karte hain): $N\sin\theta + \underbrace{\mu N\cos\theta}_{\text{friction ka inward-part}} = \frac{mv_{max}^2}{r} \tag{4}$

(3) ko $N (cos θ - μ sin θ) = m g$ se group karo, phir (4) ko isse divide karo — phir se $N$ aur $m$ gayab ho jaate hain:

PICTURE. Fast car bank par upar: $N$ lean in kar raha hai, friction arrow slope se neeche point kar raha hai, dono inward total ko feed kar rahe hain.

Step 8 — Zyada slow jaao: friction UPAR flip ho jaata hai

KYA. Agar car design speed se zyada slow chalti hai, toh gravity ka slope se neeche pull jeet jaata hai aur car inward-and-down slide karne ki koshish karti hai. Friction slope se upar flip ho jaata hai, use bahar hold karta hua — ek lowest safe speed $v_{min}$ set karta hua.

KYU. Same rule: friction slip direction ko oppose karta hai. Ab slip inward-down hai, toh friction outward-up grip karta hai. Mathematically hum sirf pichle result mein $μ$ ka sign flip karte hain:

PICTURE. Slow car bank par neeche: friction arrow ab slope se upar point kar raha hai, car ko gravity ke inward slide ke khilaaf prop kar raha hai.

Recall Do edge cases khud check karo

Jab road flat ho, $θ = 0$ , toh $v_{ma x}$ kya hai? ::: $r g \cdot μ = μ r g$ — ordinary flat-road friction limit. Formula consistent rehta hai. $v_{ma x}$ ka kya hoga jab $1 - μ tan θ \to 0$ (yaani $tan θ \to 1/ μ$ )? ::: Denominator vanish ho jaata hai, $v_{ma x} \to \infty$ — itna steep bank ki friction kisi bhi speed ko hold kar sake.

Ek-picture summary

Har force, har component, dono friction directions, aur resulting formulas — ek single cross-section mein compress kiye gaye.

Recall Feynman retelling — walkthrough simple words mein

Ek turning car har instant inward yank ho rahi hai, warna woh road se ud jaati hai (Step 1). Flat ground par sirf tyres ki grip hi yank kar sakti hai, aur grip ice par fail ho jaati hai (Step 2). Toh hum road ko ek bowl ke inside ki tarah tilt karte hain. Ab ground ka push, jo hamesha surface se seedha bahar hota hai, centre ki taraf lean karta hai aur hamare liye yanking karta hai (Step 3). Hum us tilted push ko ek up-part aur ek sideways-part mein slice karte hain (Step 4): up-part car ka weight hold karta hai, sideways-part inward yank hai (Step 5). Un dono facts ko divide karne se mass aur push ki size wipe out ho jaati hai, ek pure relation bachta hai tilt, speed, aur turn ke beech: $tan θ = v^{2} / r g$ (Step 6). Real friction add karo aur tumhe wiggle room milta hai — fast jao aur friction help karne ke liye neeche slide karta hai (Step 7); slow jao aur yeh tumhe hold karne ke liye upar flip karta hai (Step 8). Yeh ek poora road-engineering rulebook hai ek right triangle se.

Connections

Centripetal force — woh inward force jo Steps 1–2 demand karte hain.
Uniform Circular Motion — jahan se $v^{2} / r$ aata hai.
Friction — Steps 7–8 mein $μ$ , safe-speed range.
Inclined plane — same resolve-into-components method.
Conical pendulum — twin geometry, same $tan θ = v^{2} / r g$ .
Newton's Second Law — poore derivation mein har axis par apply kiya gaya.