Banking of roads — derivation

1.2.17 · Physics › Newton's Laws & Dynamics

KYA derive kar rahe hain hum

Hum banking angle $\theta$, speed $v$, radius $r$, aur friction coefficient $\mu$ ke beech ka relationship chahte hain, taaki car skid kiye bina turn kare.

KAISE: Free-body setup (car pe forces)

Banked surface pe car pe teen forces kaam karti hain:

• Weight $mg$ — seedha neeche.
• Normal force $N$ — road surface ke perpendicular (vertical se $\theta$ se tilted).
• Friction $f$ — surface ke along (slope pe upar ya neeche point kar sakti hai).

Car ek horizontal circle mein move karti hai, isliye net force horizontal aur centre ki taraf honi chahiye. Koi vertical acceleration nahi hai.

Case 1 — Frictionless banking (clean derivation)

$f=0$ set karo. $N$ ko components mein resolve karo.

Vertical & horizontal mein resolve kyun? Kyunki acceleration purely horizontal (centripetal) hai aur vertical zero — yeh natural axes hain.

Vertical equilibrium (koi vertical acceleration nahi): N\cos\theta = mg \tag{1} Yeh step kyun? $N$ ka upward vertical part gravity ko balance karta hai.

Horizontal (centripetal force deta hai): N\sin\theta = \frac{mv^2}{r} \tag{2} Yeh step kyun? $N$ ka inward horizontal part HI centripetal force hai.

$N$ aur $m$ eliminate karne ke liye (2) ko (1) se divide karo: $\frac{N\sin \theta }{N\cos \theta }=\frac{m{v}^2/r}{mg}$

Case 2 — Friction ke saath (poori picture)

Friction ek range of safe speeds ke liye adjust karta hai, sirf ek ke liye nahi.

Maximum safe speed (car UP/outward slide karne ki koshish karti hai → friction DOWN the slope kaam karta hai)

$f=\mu N$ slope ke neeche kaam karta hai.

Resolve karo. Vertical: N\cos\theta = mg + f\sin\theta = mg + \mu N\sin\theta \tag{3} Kyun? Weight neeche kheenchta hai; down-slope friction ka vertical component bhi neeche kheenchta hai.

Horizontal (centre ki taraf): N\sin\theta + f\cos\theta = \frac{mv^2}{r} \implies N\sin\theta + \mu N\cos\theta = \frac{mv_{max}^2}{r} \tag{4} Kyun? $N$ ka inward part aur friction ka inward part dono milke centripetal force banate hain.

(3) se: $N(\cos \theta -\mu \sin \theta )=mg$. (4) mein substitute karo aur divide karo: $\frac{v_{max}^2}{rg}=\frac{\sin \theta +\mu \cos \theta }{\cos \theta -\mu \sin \theta }=\frac{\tan \theta +\mu }{1-\mu \tan \theta }$

Minimum safe speed (car DOWN/inward slide karne ki koshish karti hai → friction UP the slope kaam karta hai)

$\mu$ ka sign flip karo (friction ab slope pe upar hai): $\boxed{v_{min}=\sqrt{rg\frac{\tan \theta -\mu }{1+\mu \tan \theta }}}$

(Agar $\tan \theta \le \mu$, toh $v_{min}=0$ — car bina andar slide kiye khadi bhi reh sakti hai.)

Common mistakes (Steel-manned)

Flashcards

Connections

• Centripetal force — woh force jo banking supply karne ke liye design ki gayi hai.
• Uniform Circular Motion — $v^2/r$ requirement ka source.
• Friction — $v_{min}$–$v_{max}$ range set karta hai.
• Inclined plane — same force-resolution technique.
• Conical pendulum — identical geometry ($\tan \theta =v^2/rg$).
• Newton's Second Law — $\sum F=ma$ har axis pe apply hota hai.

Concept Map