Banking of roads — derivation: Worked examples — Hinglish deep dive

1.2.17 · D3 · Physics › Newton's Laws & Dynamics › Banking of roads — derivation

Yeh child page parent derivation ko kaam mein laati hai. Hum formulas dobara derive nahi karenge — hum unhe use karenge har us situation par jo yeh topic throw kar sakta hai. Numbers touch karne se pehle, poora battlefield map karte hain taaki koi bhi scenario aapko surprise na kare.

Recall Teen formulas jo hum baar baar use karenge (parent se)

Design (frictionless): $tan θ = \frac{v ^{2}}{r g}$ , isliye $v_{0} = r g tan θ$ . Maximum with friction: $v_{ma x} = r g \frac{μ + tan θ}{1 - μ tan θ}$ . Minimum with friction: $v_{min} = r g \frac{tan θ - μ}{1 + μ tan θ}$ . Yahan $g = 9.8 m/s^{2}$ throughout hai jab tak problem kuch aur na kahe.

The scenario matrix

Har banking problem inhi cells mein se kisi ek mein hoti hai. Har cell ek alag physical question poochhti hai, isliye har ek ka apna worked example hai.

Cell	Isme kya khaas hai	Physical question
A. Pure design	$μ = 0$ , ek perfect speed poochhi	$v_{0}$ kya hai?
B. Max with friction	$μ > 0$ , bahut tez ja rahe hain	Kitni tez speed se pehle car bahar ud jaayegi?
C. Min with friction	$μ > 0$ , bahut dheere ja rahe hain	Kitni dheemi speed par car andar slip karegi?
D. Degenerate: flat road	$θ = 0$	Kya formula friction limit par collapse ho jaata hai?
E. Degenerate: $tan θ \leq μ$	bahut steep YA bahut rough	Kya car khadi reh sakti hai ( $v_{min} = 0$ )?
F. Limiting: $μ tan θ \to 1$	denominator $\to 0$	$v_{ma x} \to \infty$ kyun hota hai?
G. Inverse problem	$v$ diya, $θ$ nikalna hai	Ek target speed ke liye road design karo
H. Real-world word problem	km/h, real curve	Kya posted limit safe hai?
I. Exam twist	friction direction ambiguous	$f$ kis taraf point karta hai, aur kyun?

Ab hum saare nau cells ko 9 examples se cover karenge.

Cell A — Pure design speed

Worked example A: ek perfect speed

$r = 80 m$ radius ka ek curve $θ = 2 5^{\circ}$ par banked hai. Wahan barf hai — frictionless maano. Kaunsi ek speed par car safely turn kar sakti hai?

Forecast: Zyada steep ya choda curve → zyada tez design speed. $2 5^{\circ}$ moderate hai, $80 m$ kaafi choda hai. Andaza: kahin $20 m/s$ ke aaspaas?

$v_{0} = r g tan θ$ use karo. Yeh step kyun? $μ = 0$ ke saath bilkul ek hi safe speed hoti hai — design speed — kyunki friction ab koi bhi mismatch absorb nahi kar sakta.
$tan 2 5^{\circ} = 0.4663$ . Kyun? $tan$ = ( $N$ ka andar horizontal part) ÷ ( $N$ ka upar vertical part); yeh woh ratio hai jo tilt ko force balance se jodata hai.
$v_{0} = 80 \times 9.8 \times 0.4663 = 365.6 = 19.12 m/s$ .

Verify: Units: $m \cdot (m/s^{2}) = m^{2} / s^{2} = m/s$ ✓. Value $\approx 19.1 m/s \approx 69 km/h$ — forecast ke band se match karta hai. ✓

Cell B — Maximum speed with friction

Worked example B: kitni tez speed par car bahar ud jaayegi

Wohi curve $r = 80 m$ , $θ = 2 5^{\circ}$ , lekin ab dry road $μ = 0.35$ . Maximum safe speed kya hai?

Forecast: Friction ab car ko andar rakhne mein madad karta hai, isliye $v_{ma x}$ design $19.1 m/s$ se zyada hona chahiye.

$v_{ma x} = r g \frac{μ + tan θ}{1 - μ tan θ}$ use karo. Yeh step kyun? Design speed se zyada par car upar/bahar slide karne ki koshish karti hai, isliye friction slope ke neeche point karta hai aur apna inward component add karta hai — yahi " $+ μ$ on top" formula hai.
Top: $μ + tan θ = 0.35 + 0.4663 = 0.8163$ .
Bottom: $1 - μ tan θ = 1 - 0.35 \times 0.4663 = 1 - 0.1632 = 0.8368$ . Kyun? Down-slope friction ka ek chhota downward vertical part bhi hota hai, jo $N$ ko reduce karta hai — isse denominator chota ho jaata hai.
Ratio $= 0.8163/0.8368 = 0.9755$ .
$v_{ma x} = 80 \times 9.8 \times 0.9755 = 764.8 = 27.66 m/s$ .

Verify: $27.7 > 19.1$ ✓ (friction ne ceiling badhaa di). Units clean hain. ✓

Cell C — Minimum speed with friction

Worked example C: kitni dheemi speed par car andar slip karegi

Wohi curve, wohi $μ = 0.35$ . Minimum kitni speed par car andar slide nahi karegi?

Forecast: Design speed se kam par car neeche/andar slip karne ki koshish karti hai, isliye friction ab iske against kaam karta hai, slope ke upar point karta hai. $v_{min}$ ko $19.1 m/s$ se kam hona chahiye.

$v_{min} = r g \frac{tan θ - μ}{1 + μ tan θ}$ use karo. Yeh step kyun? Friction ne direction reverse kar li (up-slope) → max formula mein $μ$ ka sign flip karo.
Top: $tan θ - μ = 0.4663 - 0.35 = 0.1163$ . Sign check kyun? Agar yeh negative ho jaata, toh $v_{min}$ imaginary hota → matlab $0$ (Cell E dekho).
Bottom: $1 + μ tan θ = 1.1632$ .
Ratio $= 0.1163/1.1632 = 0.09998$ .
$v_{min} = 80 \times 9.8 \times 0.09998 = 78.4 = 8.85 m/s$ .

Verify: $8.85 < 19.1 < 27.7$ — safe band $[v_{min}, v_{ma x}]$ sahi tarike se design speed ko beech mein rakhta hai. ✓

Cell D — Degenerate: flat road ( $θ = 0$ )

Cell E — Degenerate: $tan θ \leq μ$ (car khadi reh sakti hai)

Worked example E: steep + rough →

$v_{min} = 0$ Ek steep bobsled-style curve, $r = 30 m$ , $θ = 4 0^{\circ}$ , rough surface $μ = 0.9$ . $v_{min}$ nikalo.

Forecast: Bahut rough aur bahut steep. Shayad friction akela car ko jagah par rok sake, isliye $v_{min}$ $0$ ho sakta hai.

$v_{min}$ ka top check karo: $tan 4 0^{\circ} - μ = 0.8391 - 0.9 = - 0.0609$ . Pehle check kyun? Negative numerator matlab square root imaginary hogi — physically real speed ke liye impossible.
Kyunki $tan θ \leq μ$ hai, car ko rest par bhi andar slide karne ki koi tendency nahi — friction enough up-slope hold provide kar sakta hai. Isliye hum $v_{min} = 0$ set karte hain. Kyun? Formula ka "negative inside root" nature ka tarika hai yeh kehne ka ki "yeh case ho nahi sakta; sach mein minimum khadi rehna hai."

Verify: $tan 4 0^{\circ} = 0.8391 \leq 0.9 = μ$ ✓ standstill condition confirm karta hai. Isliye $v_{min} = 0 m/s$ . ✓

Cell F — Limiting: $μ tan θ \to 1$ (denominator vanish ho jaata hai)

Worked example F:

$v_{ma x}$ kyun blow up karta hai Ek curve $r = 30 m$ , $θ = 4 5^{\circ}$ ( $tan θ = 1$ ), $μ = 1.0$ ke saath. $v_{ma x}$ kya karta hai?

Forecast: Denominator $1 - μ tan θ = 1 - 1 = 0$ . Zero se divide karna → $v_{ma x} \to \infty$ . Kuch dramatic hoga.

Denominator $= 1 - μ tan θ = 1 - 1 \times 1 = 0$ . Yeh kyun matter karta hai: down-slope friction ki grip $N$ ke saath itni tezi se badhti hai ki koi bhi finite speed ise overpower nahi kar sakti — car theoretically kabhi bahar nahi jaati.
Jaise jaise $μ tan θ \to 1^{-}$ , $\frac{μ + tan θ}{1 - μ tan θ} \to + \infty$ , isliye $v_{ma x} \to \infty$ . Physically kyun? Zyada speed ke liye zyada $N$ chahiye, lekin zyada $N$ matlab zyada friction, jo yahan andar point karta hai — ek self-reinforcing loop jiska koi upper break nahi.

Verify: $μ = 0.999$ , $tan θ = 1$ ke saath: ratio $= \frac{1.999}{0.001} = 1999$ , jo $v_{ma x} = 30 \times 9.8 \times 1999 \approx 766 m/s$ deta hai — bahut bada, blow-up trend confirm karta hai. ✓

Cell G — Inverse problem (road design karna)

Cell H — Real-world word problem

Cell I — Exam twist: friction kis taraf point karta hai?

Yeh classic trap hai. Friction ki direction fixed nahi hoti — yeh depend karta hai ki given speed design speed se upar hai ya neeche. Figure dekho.

Worked example I: friction ki direction decide karo, phir solve karo

Ek curve mein $r = 100 m$ , $θ = 2 0^{\circ}$ , $μ = 0.30$ hai. Car $v = 30 m/s$ par chal rahi hai. Friction kis taraf point karta hai, aur kya car safe hai?

Forecast: Pehle design speed nikalo. Agar $30 m/s$ isse upar hai, toh car bahar ki taraf tend karti hai → friction down-slope; max comparison use karo.

Design speed: $v_{0} = r g tan 2 0^{\circ} = 100 \times 9.8 \times 0.36397 = 356.7 = 18.89 m/s$ . Pehle kyun? Yahi pivot hai: isse neeche friction upar point karta hai, isse upar friction neeche point karta hai.
Kyunki $30 > 18.89$ , car upar/bahar slide karne ki koshish karti hai, isliye friction slope ke neeche act karta hai (figure mein red arrow). Kyun? Friction relative sliding ki tendency ko oppose karta hai, gravity ko nahi.
Ceiling check karo: $v_{ma x} = r g \frac{μ + tan θ}{1 - μ tan θ} = 980 \frac{0.30 + 0.36397}{1 - 0.30 \times 0.36397}$ .
Ratio $= \frac{0.66397}{0.89081} = 0.74535$ ; $v_{ma x} = 980 \times 0.74535 = 730.4 = 27.03 m/s$ .

Verify: $30 m/s > 27.03 m/s = v_{ma x}$ → car maximum exceed karti hai aur bahar skid kar jaayegi. Unsafe. Direction call (down-slope) confirm hoti hai kyunki $v > v_{0}$ . ✓

Recall check

Friction kis taraf point karta hai jab

v > v_{0}

Slope ke neeche (car bahar/upar slide karne ki koshish karti hai).

Friction kis taraf point karta hai jab

v < v_{0}

Slope ke upar (car andar/neeche slip karne ki koshish karti hai).

v_{min} = 0

kab hota hai?

Jab

tan θ \leq μ

— friction akela car ko rest par bhi rok sakta hai.

v_{ma x} \to \infty

kyun hota hai jab

μ tan θ \to 1

Denominator

1 - μ tan θ \to 0

; friction ki inward grip

N

ke saath itni tezi se badhti hai ki koi bhi speed ise overpower nahi kar sakti.

Target design speed ke liye banking angle kaise nikaalte hain?

θ = arctan \frac{v ^{2}}{r g}

Connections

Centripetal force — har example mein $m v^{2} / r$ supply karne ke liye forces balance kiye jaate hain.
Friction — cells B, C, E, F, I mein $v_{min}$ – $v_{ma x}$ range set karta hai.
Uniform Circular Motion — $v^{2} / r$ requirement ka origin.
Inclined plane — wohi resolve-into-axes method yahan bhi use hoti hai.
Conical pendulum — design formula $tan θ = v^{2} / r g$ share karta hai.
Newton's Second Law — har solution mein per axis apply hota hai.
Parent derivation par wapas jaao: parent derivation.