Cars on level and banked roads

Turning on a level road

Narrated video. Transcript:

A car goes round a curve at steady speed. Its direction keeps changing, so it is accelerating towards the centre of the curve.

Acceleration needs a net force. For a body moving in a circle, that force must point to the centre and have size m v squared over r.

On a level road, what supplies it? Draw the forces on the car: weight m g down, normal force N up. They cancel, so N equals m g.

The only sideways force is friction between the tyres and the road. It acts at the tyres, pointing towards the centre.

It is static friction, because the tyres do not slide sideways. So it has a limit: f is at most mu s times N, which is mu s m g.

For the car to turn, m v squared over r must not exceed mu s m g. The mass cancels, and v squared must be at most mu s r g.

So the maximum safe speed on a level curve is the square root of mu s r g. It does not depend on the mass of the car.

Go faster and friction cannot hold: the car skids outward. On a wet road mu s is smaller, so the safe speed drops. Slow down in the rain!

Why roads are banked

Narrated video. Transcript:

On highway curves and racetracks, the outer edge of the road is raised. The road is tilted at an angle theta. This is called banking.

Now the normal force N is perpendicular to the tilted road, so it leans towards the centre. Part of N, N sin theta, pushes the car round the curve.

At high speed, friction acts down the slope. Vertically: N cos theta equals m g plus f sin theta. Towards the centre: N sin theta plus f cos theta equals m v squared over r.

Put in f at its limit, mu s N, and solve. The maximum speed is the square root of r g times mu s plus tan theta, over one minus mu s tan theta.

Now take away friction completely: mu s equals zero. There is still one speed that works: v nought equals the square root of r g tan theta.

At this optimum speed the normal force alone turns the car. No friction is needed, so there is the least wear and tear on the tyres.

Drive slower than v nought and the car tends to slide down the bank, so friction acts up the slope. Drive faster and friction acts down the slope.

Example: a racetrack of radius 300 metres banked at 15 degrees, with mu s equal to 0.2. The optimum speed is 28.1 metres per second, and the maximum is 38.1 metres per second.

Try it: a car on a curve

Interactive simulation. Things to try:

• Drag to orbit the view, or press Driver view to ride behind the car.
• Speed up until the car skids outward. Compare with the maximum safe speed readout.
• Press Drive at v₀: the friction arrow disappears because no friction is needed.
• Set the banking angle to 0° for a level road. Now friction alone supplies the force towards the centre.

Quick check

A cyclist rides at 18 km/h round a level circular track of radius 3 m. The coefficient of static friction between the tyres and the road is 0.1. Will the cyclist slip?

1. No, because the speed is low
2. Yes, because v² is greater than μs r g
3. No, because friction always equals mv²/r
4. It depends on the mass of the cyclist

The banked racetrack

A racetrack of radius 300 m is banked at 15°. The coefficient of static friction between the tyres and the road is 0.2. What is the maximum speed before the car skids? (g = 9.8 m/s², tan 15° ≈ 0.27)

1. 24.2 m/s
2. 28.1 m/s
3. 38.1 m/s
4. 54.2 m/s

Summary

• Level road: vₘₐₓ = √(μs r g)
• Banked road: vₘₐₓ = √[rg(μs + tan θ) / (1 − μs tan θ)]
• Optimum speed (no friction needed): v₀ = √(rg tan θ)

• A car on a curve needs a net force mv²/r towards the centre.
• On a level road only static friction supplies it, so the safe speed is limited to √(μs r g).
• On a banked road the normal force also leans inwards and helps, so higher speeds are safe.
• At the optimum speed v₀ no friction is needed. Slower than v₀, friction acts up the slope; faster, it acts down the slope.