Banking & Circular Motion
Take a car around a flat or banked curve and find the fastest safe speed.
About this simulation
Drive a car round a curved road. Tilt the road, change how grippy it is and how fast the car goes, and see whether the car holds the turn, skids outward or slides down the slope. Arrows show every force on the car, and a view from above shows it going round. A speed gauge shows how close you are to skidding, and the forces drawn tip to tail show the gap when you do. Switch to a cyclist, who leans instead of the road, or to the conical pendulum, a ball whirling on a string: the same physics each time.
Try it and see
-
Drag Speed slowly from 0 to 40 m/s.
Friction points up the slope below 13.4 m/s, vanishes at 13.4 (v₀), and points down the slope above it. At 19.1 m/s it hits f_max and the car skids outward.
-
Tap 'Flat road', then raise Speed.
With no bank, friction alone turns the car. At 10 m/s the grip meter reads 68%; past v_max = √(μrg) = 12.1 m/s it runs out and the car skids.
-
Tap 'Frictionless bank', then move Speed one step either way.
No friction arrow: the bank alone turns the car, N sinθ = mv²/r, at v₀ = 11.0 m/s. At 11.5 m/s it slides up and out; at 10.5 m/s it slides down. Without friction only one speed works.
-
Tap 'Wet highway bend', then raise Friction to 0.50.
Wet, v_max is 96 km/h and 28 m/s (101 km/h) needs 113% of the grip there is: the forces tip to tail fall short of mv²/r, and that gap is the skid. Dry, the same bend holds at 57%.
-
Tap 'Cyclist leans into a bend', then raise Speed.
The rider leans until N + f runs through the centre of mass: tanα = v²/rg, the bank angle's formula. Past v_max = √(μrg) the tyres cannot give the f that lean needs, and they slide out.
The physics, step by step
A car of mass m drives at a steady speed v round a horizontal circle of radius r, on a road tilted at angle θ. The road rises away from the centre. Here is how every formula on this page follows from one free-body diagram.
-
Draw the free-body diagram
Only three forces act on the car: its weight mg, straight down; the normal force N from the road, perpendicular to the road surface; and friction f, along the road surface.
There is no separate 'centripetal force'. mv²/r is what these three must add up to for the car to go round the circle.
-
Choose horizontal and vertical axes
The car moves in a horizontal circle, so its acceleration v²/r points horizontally toward the centre, not along the slope. Taking axes along and across the slope is the usual mistake: then the acceleration has parts on both axes. With horizontal and vertical axes, the acceleration lies entirely on one of them.
-
Resolve each force
N is tilted from the vertical by θ, the same angle as the road. So N has a part N sinθ toward the centre and a part N cos θ upward. Take friction acting down the slope, which is its direction when the car is going fast. Then it has a part f cosθ toward the centre and a part f sinθ downward. Weight is already vertical.
-
Apply Newton's second law on each axis
Nothing moves vertically, so the vertical forces balance. Horizontally the net force equals mass times the centripetal acceleration:
Vertical: N cosθ − f sinθ − mg = 0Horizontal: N sinθ + f cosθ = mv²/r -
No friction: the ideal speed v₀
Put f = 0. Then N cosθ = mg and N sinθ = mv²/r. Divide the second by the first: N and m cancel.
tanθ = v₀² / rgv₀ = √(rg tanθ) -
Fastest safe speed: v_max
The faster the car, the more friction must help, down the slope. Friction can be at most f_max = μN. At v_max it is exactly that, so put f = μN:
Vertical: N (cosθ − μ sinθ) = mgHorizontal: N (sinθ + μ cosθ) = mv²/rDivide: v²/rg = (sinθ + μ cosθ) / (cosθ − μ sinθ)Divide top and bottom by cosθ:v_max = √( rg (tanθ + μ) / (1 − μ tanθ) )At exactly v_max both equations still hold. Any faster and friction cannot grow past f_max, so neither balances: the sideways sum falls short of mv²/r and the vertical sum N cosθ − f sinθ is now bigger than mg. The car slides up the bank and out of the circle, rising as it goes. The simulation shows both numbers when it skids.
-
Slowest safe speed: v_min
Too slow, and the car tends to slide down the bank, so friction acts up the slope. The same working with f = μN up the slope flips the signs:
v_min = √( rg (tanθ − μ) / (1 + μ tanθ) ) -
Special cases
If tanθ ≤ μ, the top of the v_min fraction is zero or negative, so there is no minimum: the car could stop on the bank without sliding. On a flat road, θ = 0, and everything comes from friction. m cancels in every result, so the safe speeds do not depend on the car's mass.
Flat road: v_max = √(μ r g) -
The same maths with a string: the conical pendulum
A ball on a string of length L whirls in a horizontal circle with the string at angle φ to the vertical. Tension T takes the place of N, and there is no friction. The circle's radius is r = L sinφ.
Vertical: T cosφ = mgHorizontal: T sinφ = mv²/rtanφ = v² / rg (the frictionless bank, again)Period: T = 2π √(L cosφ / g) -
A cyclist leans instead of the road
On a flat road the ground pushes up on the tyre with N = mg and sideways with friction f = mv²/r. Together they make one force from the tyre. If its line misses the centre of mass it turns the rider over, so the rider leans until it runs through: the lean angle α from the vertical is the angle of that force.
tanα = f / N = v² / rg (the frictionless bank, again)Grip limit: f ≤ μmg → v_max = √(μ r g), α_max = tan⁻¹ μ
Questions
Why are roads banked on curves?
A car going round a curve needs a force towards the centre. On a flat road only friction between the tyres and the road can supply it, and friction is limited and unreliable on wet roads. Tilting the road makes the normal force lean towards the centre, so it provides part of the centripetal force. This raises the safe speed and reduces wear on the tyres. In the simulation, compare 'Flat road' with a banked one.
What is the maximum safe speed on a flat circular road?
On a flat road the only horizontal force is friction, which can be at most μmg. Setting μmg = mv²/r gives v_max = √(μrg). Mass cancels, so a heavy truck and a light car have the same limit on the same road. In the simulation, set the bank angle to 0 and check that v_max on the speed gauge equals √(μrg).
What is the optimum speed on a banked road?
The optimum or ideal speed is the speed at which the normal force alone provides the centripetal force, so no friction is needed. Resolving forces gives N cos θ = mg and N sin θ = mv²/r, so tan θ = v²/rg and v = √(rg tan θ). Tyres wear least at this speed. The 'ideal' figure under the simulation's verdict shows this value for the angle and radius you choose.
Can a car slip down a banked road if it goes too slowly?
Yes. Below the ideal speed, the slope pulls the car inward more than the turn requires, so friction must act up the slope to hold it. If the speed is below the minimum safe speed, friction cannot do enough and the car slides down towards the inside of the curve. In the simulation, lower the speed on a steep bank with little friction to see 'Slips inward'.
How is a conical pendulum similar to a banked road?
In a conical pendulum the bob moves in a horizontal circle while the string makes an angle φ with the vertical. The vertical part of the tension balances the weight and the horizontal part provides the centripetal force, giving tan φ = v²/rg, the same form as the banked road. Its period is T = 2π√(L cos φ/g). The simulation's Conical pendulum set-up draws the bob's circle and labels the string length L and the angle.
Why does a cyclist lean when turning?
Friction from the road turns the bike, but it acts at the tyre, below the rider's centre of mass. Upright, it would tip the rider outward. Leaning by α with tan α = v²/rg makes the road's total push (N and friction together) pass through the centre of mass, so it turns the rider without tipping them. In the simulation, choose the Cyclist set-up and raise the speed: the lean grows, and past √(μrg) the tyres slide out.
