Banking of roads — find safe speed for car on banked curve with friction

Question

A car moves on a banked curve of radius $R = 100$ m banked at angle $\theta = 30°$ . The coefficient of friction between the tyres and the road is $\mu = 0.2$ . Find the maximum safe speed of the car.

(JEE Main 2022, similar pattern)

Solution — Step by Step

Three forces act: weight $mg$ downward, normal reaction $N$ perpendicular to the banked surface, and friction $f$ along the surface. At maximum speed, the car tends to slide outward (up the incline), so friction acts inward (down the incline).

Vertical equilibrium (no vertical motion):

N\cos\theta - f\sin\theta = mg

Horizontal equation (centripetal force):

N\sin\theta + f\cos\theta = \frac{mv^2}{R}

At the maximum speed, friction is at its limit: $f = \mu N$ .

From the vertical equation: $N(\cos\theta - \mu\sin\theta) = mg$

From the horizontal equation: $N(\sin\theta + \mu\cos\theta) = \frac{mv^2}{R}$

Divide the second by the first:

\frac{v^2}{Rg} = \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta} = \frac{\tan\theta + \mu}{1 - \mu\tan\theta}

v_{max} = \sqrt{Rg \cdot \frac{\tan\theta + \mu}{1 - \mu\tan\theta}}

$\tan 30° = 1/\sqrt{3} \approx 0.577$

v_{max} = \sqrt{100 \times 10 \times \frac{0.577 + 0.2}{1 - 0.2 \times 0.577}}

= \sqrt{1000 \times \frac{0.777}{0.8846}} = \sqrt{1000 \times 0.878}

\boxed{v_{max} \approx 29.6 \text{ m/s} \approx 107 \text{ km/h}}

Why This Works

Banking tilts the normal force so that a component of $N$ itself provides centripetal force — even without friction. Friction gives an extra “push” inward at higher speeds (or acts outward at very low speeds to prevent the car from sliding down).

The formula $v_{max} = \sqrt{Rg(\tan\theta + \mu)/(1 - \mu\tan\theta)}$ elegantly captures both effects. Without friction ( $\mu = 0$ ), it reduces to $v = \sqrt{Rg\tan\theta}$ — the ideal banking speed.

Alternative Method

For a quick check, note that the expression has the same structure as the angle addition formula for tangent: $\tan(\theta + \phi)$ where $\tan\phi = \mu$ . So $v_{max}^2 = Rg\tan(\theta + \phi)$ . This is the “friction angle” approach — fast and elegant.

For the minimum speed (below which the car slides inward), flip the sign of $\mu$ :

v_{min} = \sqrt{Rg \cdot \frac{\tan\theta - \mu}{1 + \mu\tan\theta}}

JEE sometimes asks for both $v_{max}$ and $v_{min}$ in the same problem.

Common Mistake

The most common error: writing friction as always acting “up the incline.” At maximum speed, friction acts down the incline (inward). At minimum speed, friction acts up the incline (outward). The direction of friction depends on whether the car tends to slide outward or inward. Draw the FBD for the specific case asked — do not memorise a single direction.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next