Banked RoadJEE Advanced

Concept

On a banked road the track is tilted so the normal force gains an inward horizontal component that helps provide the centripetal force. At the ideal speed friction is not needed at all; at other speeds friction makes up the difference, giving a safe band of speeds.

Key formula

Derivation

With banking angle $\theta$ and no friction, resolve the normal force $N$: vertically $N\cos\theta = mg$, horizontally $N\sin\theta = mv^2/r$. Dividing gives $\tan\theta = v_0^2/(rg)$, so $v_0 = \sqrt{rg\tan\theta}$.

Add friction $f \le \mu N$ acting down the slope at the maximum speed. Resolving along and perpendicular and eliminating $N$ yields $v_{\max} = \sqrt{rg(\tan\theta + \mu)/(1 - \mu\tan\theta)}$.

At the minimum speed friction points up the slope, giving $v_{\min} = \sqrt{rg(\tan\theta - \mu)/(1 + \mu\tan\theta)}$, which is real only when $\tan\theta > \mu$.

Scenarios to explore

• Banked Road — Optimum speed on a banked curve with friction.
• Ideal banking: no friction needed — At the design speed v = √(rg tanθ) the car stays on the road even on ice.
• Racetrack banking — A steep bank with grippy tyres allows very high speeds.
• Icy bend — Gentle bank, almost no friction: the safe speed band collapses.

Real-world applications

• Highway and race-track curve design.
• Railway tracks super-elevated on bends.
• Velodromes for cycling.

JEE exam tips

• At the ideal speed, set friction to zero — the algebra is much cleaner.
• Convert m/s to km/h with ×3.6 when a problem quotes road speeds.

Common mistakes

• Assuming friction always acts down the slope — its direction flips between $v_{\min}$ and $v_{\max}$.
• Forgetting the $mg$ contribution when resolving forces perpendicular to the incline.
• Treating the ideal speed as the maximum speed.

Exam traps to avoid

• If $\mu\tan\theta \ge 1$ there is no finite maximum speed — the car cannot skid outward.
• The mass cancels everywhere; safe speeds are independent of the vehicle's mass.