A car on a banked road of radius 50m — find safe speed for 30° banking

Question

A car moves on a banked road of radius 50 m. The banking angle is 30°. Find the optimum speed at which the car can negotiate the bend without relying on friction. (Take $g = 10$ m/s²)

Solution — Step by Step

On a banked road, two forces act on the car:

Weight $mg$ (vertically downward)
Normal force $N$ (perpendicular to the banked surface)

At the optimum speed, friction is zero. The normal force and weight together provide the centripetal force.

We resolve $N$ into components:

Vertical component: $N\cos\theta$ (balances weight)
Horizontal component: $N\sin\theta$ (provides centripetal force)

Vertical equilibrium (no vertical acceleration):

N\cos\theta = mg \quad \text{...(1)}

Horizontal (centripetal) direction:

N\sin\theta = \frac{mv^2}{r} \quad \text{...(2)}

Dividing equation (2) by equation (1):

\frac{N\sin\theta}{N\cos\theta} = \frac{mv^2/r}{mg}

\tan\theta = \frac{v^2}{rg}

v^2 = rg\tan\theta

v^2 = rg\tan\theta = 50 \times 10 \times \tan 30°

\tan 30° = \frac{1}{\sqrt{3}} \approx 0.577

v^2 = 500 \times \frac{1}{\sqrt{3}} = \frac{500}{\sqrt{3}} \approx 288.7 \text{ m}^2/\text{s}^2

v = \sqrt{288.7} \approx 17.0 \text{ m/s} \approx 61 \text{ km/h}

\boxed{v_{optimum} = \sqrt{\frac{rg}{\sqrt{3}}} = \sqrt{\frac{500}{\sqrt{3}}} \approx 17 \text{ m/s}}

Why This Works

Banking a road redirects the normal force. On a flat road, the normal force is purely vertical (balancing gravity) and friction must provide all centripetal force. On a banked road, even without friction, the horizontal component of $N$ provides centripetal force.

The formula $\tan\theta = v^2/(rg)$ is derived by requiring these two conditions simultaneously: vertical equilibrium and centripetal acceleration. The elegant result is that at this specific speed $v = \sqrt{rg\tan\theta}$ , the car needs zero friction. This is why highway engineers bank curves — it’s designed for a specific design speed.

At speeds above this optimum, friction acts inward (down the slope) to provide extra centripetal force. At speeds below, friction acts outward (up the slope) to prevent the car from sliding inward. Friction gives the car a range of safe speeds, not just one.

Alternative Method

Directly memorise the formula: $v_{optimum} = \sqrt{rg\tan\theta}$ .

For this problem: $v = \sqrt{50 \times 10 \times \tan 30°} = \sqrt{500/\sqrt{3}} \approx 17$ m/s.

The derivation is important to understand, but in an exam, deriving from scratch every time wastes 3–4 minutes.

Common Mistake

A frequent mistake is using $\sin\theta$ instead of $\tan\theta$ in the formula, i.e., writing $v^2 = rg\sin\theta$ . This comes from forgetting to divide the two equations and instead using just one component equation. Always divide the two equations to get $\tan\theta$ — the $N$ and $m$ cancel neatly, which is why the formula has no $m$ (the optimum speed is mass-independent).

A related formula for the maximum and minimum safe speeds with friction coefficient $\mu$ on a banked road:

$v_{max} = \sqrt{rg \cdot \frac{\tan\theta + \mu}{1 - \mu\tan\theta}}$ , $\quad v_{min} = \sqrt{rg \cdot \frac{\tan\theta - \mu}{1 + \mu\tan\theta}}$$

These appear in JEE. The middle value $v_{opt} = \sqrt{rg\tan\theta}$ (zero friction case) lies between them.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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