Banked Curve: Friction-Free Turning Angle

A banked curve is tilted inward so that the horizontal component of the normal force provides the centripetal force needed for the car to turn — without relying on friction between tires and road. Race tracks, highways, and railroad curves all use banking to make turns safer and faster.

The Physics

For a car of mass $m$ going around a curve of radius $r$ at speed $v$ on a banked surface tilted at angle $\theta$:

• The normal force $N$ is perpendicular to the road surface, so it has a horizontal component $N\sin\theta$ pointing inward (toward the center of the curve) and a vertical component $N\cos\theta$ pointing up.
• For circular motion: the horizontal component of $N$ supplies the centripetal force.
• For vertical equilibrium: the vertical component of $N$ balances gravity.

$N\sin\theta = \dfrac{mv^{2}}{r} \qquad N\cos\theta = mg$

Divide the first equation by the second to eliminate both $N$ and $m$:

$\tan\theta = \dfrac{v^{2}}{rg}$

The required banking angle depends only on the speed, the radius, and gravity — not on the mass of the car.

Step-by-Step Solution

Given: $r = 50\;\text{m}$, $v = 20\;\text{m/s}$, $g = 9.81\;\text{m/s}^{2}$.

Find: The banking angle $\theta$.

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Step 1 — Compute $v^{2}/(rg)$.

$\dfrac{v^{2}}{rg} = \dfrac{(20)^{2}}{50 \times 9.81}$

$= \dfrac{400}{490.5}$

$\approx 0.8155$

Step 2 — Take the inverse tangent.

$\theta = \arctan(0.8155)$

$\approx 39.21°$

Step 3 — Verify by computing the forces.

If the car has, say, $m = 1000\;\text{kg}$:

$N\cos\theta = mg \;\;\Longrightarrow\;\; N = \dfrac{mg}{\cos\theta} = \dfrac{1000 \times 9.81}{\cos 39.21°} = \dfrac{9810}{0.7745} \approx 12{,}666\;\text{N}$

Centripetal force needed:

$\dfrac{mv^{2}}{r} = \dfrac{1000 \times 400}{50} = 8000\;\text{N}$

Horizontal component of $N$:

$N\sin\theta = 12{,}666 \times \sin 39.21° = 12{,}666 \times 0.6322 \approx 8005\;\text{N}\;\;\checkmark$

(Off by 5 N due to rounding — exact within precision.)

Step 4 — What if the car is heavier or lighter?

The mass cancels out of $\tan\theta = v^{2}/(rg)$, so the same banking angle works for any vehicle at this speed and radius. A motorcycle and a tractor-trailer can both negotiate the same banked turn at the same speed.

Step 5 — What if the car goes faster than the design speed?

If $v$ is too high, $v^{2}/(rg) > \tan\theta_{\text{actual}}$, and the horizontal component of $N$ alone isn't enough to keep the car on the curve. The car will tend to slide outward — friction (or fear) takes over. A real banked curve has a "design speed" where friction is exactly zero, and ranges of speed above and below that work too if friction is available.

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Answer: The required banking angle for $r = 50\;\text{m}$ and $v = 20\;\text{m/s}$ is

$\boxed{\;\theta = \arctan\!\left(\dfrac{v^{2}}{rg}\right) = \arctan(0.816) \approx 39.21°\;}$

This is the angle at which the road would need to be tilted so that no friction is needed. The mass of the vehicle doesn't matter.

Try It

• Adjust the speed and radius sliders.
• Watch the road tilt to the new required angle.
• The HUD shows the live banking angle and confirms the force balance.
• Notice that doubling the speed quadruples $\tan\theta$, dramatically increasing the angle.
• For real highway interchanges (typically $\theta \approx 6$–$12°$), the design speed is around $v \approx \sqrt{rg\tan\theta}$.