Physics · Mechanics · Grade 10

Banked turn

What keeps a car on a bend, and why are roads banked?

The situation

On big roads a bend is usually banked slightly towards the outside, and on racing circuits the bends are made as steep as walls. In summer you take a bend at 90 km/h without a care, but in winter on an icy road the car does not turn and slides towards the edge of the road. What makes a car go round a circle? Why is a road built banked, and what is the greatest speed at which a bend can be taken?

Topic

Circular motion. A car on a bend: why roads are banked

Key ideas

Centripetal acceleration
Force of friction
Banked road

Formulas

a = \frac{v^{2}}{R}

Centripetal acceleration: the direction of the velocity changes continuously.

N \sin θ + F \cos θ = \frac{m v^{2}}{R}, N \cos θ - F \sin θ = m g

Newton's second law on a banked ( $θ$ ) road: horizontal and vertical components.

v_{0} = \sqrt{g R tg θ}

Design speed: at this speed no friction is needed at all.

v_{m a x} = \sqrt{g R \frac{μ + tg θ}{1 - μ tg θ}}

The greatest speed at which the bend can be taken without skidding. On a flat road $v_{m a x} = \sqrt{μ g R}$ .

v_{m i n} = \sqrt{g R \frac{tg θ - μ}{1 + μ tg θ}}

On a steep, slippery road ( $\tan θ > μ$ ), if the car goes slower than this it slides down.

Theory

Turning needs a force

In uniform motion in a circle, even if the magnitude of the velocity does not change, its direction does — so there is an acceleration: $a = v^{2} / R$ , directed towards the centre (green arrow). This acceleration is provided by the centripetal force $F = m v^{2} / R$ . On a flat road this force is provided only by the friction between the tyres and the road, so $v_{m a x} = \sqrt{μ g R}$ .

Why is a road banked?

If the road is banked at an angle $θ$ , the normal reaction force $N$ also tilts towards the centre and itself provides part of the centripetal force. At a speed of $v_{0} = \sqrt{g R \tan θ}$ it provides all of it — friction is zero, and the car turns even on ice. If you go faster, friction points down (towards the centre); if slower, up.

Skidding: along the tangent, by inertia

When the required friction exceeds $μ N$ , the tyres begin to slip. Nobody “throws” the car outwards: by inertia it tends to go on along the tangent, while the road turns away from under it. On ice the track goes almost along the tangent — watch the tangent line in the “From above” camera. On a very steep, slippery bend a slow-moving car, on the other hand, slides downwards, towards the centre.

Where the formula comes from: a = v²/R

In a short time $Δ t$ the car covers an arc $Δ s = v Δ t$ and the velocity vector turns through an angle $Δ φ = Δ s / R$ . If the magnitude of the velocity is constant, ${\vec{v}}_{1}$ , ${\vec{v}}_{2}$ and $Δ \vec{v} = {\vec{v}}_{2} - {\vec{v}}_{1}$ form an isosceles triangle: $| Δ \vec{v} | \approx v Δ φ = \frac{v^{2} Δ t}{R}$ . So $a = \frac{| Δ \vec{v} |}{Δ t} = \frac{v^{2}}{R}$ , and as $Δ t \to 0$ , $Δ \vec{v}$ points along the radius towards the centre. In terms of angular velocity $a = ω^{2} R = ω v$ . Look in the “From above” camera: the velocity arrow keeps the same length and only turns.

Normal and tangential acceleration

In curvilinear motion the acceleration has two parts: the normal (centripetal) $a_{n} = v^{2} / R$ changes the direction of the velocity, and the tangential $a_{τ} = Δ v / Δ t$ changes its magnitude; the total acceleration is $a = \sqrt{a_{n}^{2} + a_{τ}^{2}}$ . The car enters the bend at $0.6 v$ and accelerates at 3 m/s²: at that moment both green arrows are visible in the scene, and on the graph $a_{τ} = 3$ m/s² while $a_{n}$ grows in proportion to $v^{2}$ . Once the speed is reached, $a_{τ} = 0$ — uniform circular motion (for details see “Circular motion: the bicycle”).

Model simplifications

The car is a point mass; the engine keeps the speed along the road constant, and only the transverse (side) friction is taken into account. The coefficient of friction in skidding is also taken as $μ$ ; air resistance and the complex properties of tyres are ignored. After leaving the road the deceleration in the gravel zone is $0.5 g$ . At the start the car accelerates from $0.6 v$ to the chosen speed at $3$ m/s².

Did you know?

The main bends of the Daytona racing circuit in the USA are banked at 31° — cars take them at more than 300 km/h.
The bends of velodromes are very steep — about 40–45°: the cyclist rides almost on a wall and, if he slows down, slides down.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

More experiments in this section

Elasticity and deformation — Grade 9
Why doesn't a ship sink? — Grade 10
Simple machines: lever, pulley — Grade 10
Basketball: the free throw — Grade 10
Momentum and its conservation — Grade 10
Centre of gravity and equilibrium — Grade 10
Aeroplane wing — Grade 10
Dynamics of a rigid body — Grade 10
Non-inertial frames of reference — Grade 10
Inclined plane and Atwood machine — Grade 10