B. Dynamics · Dynamics: Forces and Motion in a PlaneBanked Curves: Design Speed, Friction and Highway Engineering
B2.6 B3.2 B1.1About 70 minutes
Learning goals- I can explain how banking a curve lets the normal force supply the centripetal force.
- I can derive and use tan θ = v² ÷ (rg) for the design (ideal) speed of a banked curve.
- I can decide which way friction acts on a banked curve when a vehicle goes faster or slower than the design speed, and find the speed limits.
- I can analyse how engineers choose bank angles for highways, race tracks and velodromes, including in snowy climates.
Words to know
banked curve
A curve whose road surface is tilted toward the centre of the curve.
bank angle (θ)
The angle between the road surface and the horizontal.
design speed (ideal speed)
The speed at which a vehicle can round a banked curve with no friction needed: v = √(rg tan θ).
superelevation
The highway-engineering term for banking a road on a curve.
Leaning into the turn
Cycling tracks (velodromes) have curves so steep they look like walls. Highway ramps are tilted gently toward the inside. Even a plane tilts its wings to turn. Why?
On a flat curve, only friction can push a car toward the centre, and in a Canadian winter friction is not reliable. Tilting the road lets the normal force, which is always there, lean inward. Part of it then points toward the centre and does the centripetal job, even on ice.
The design speed
On a frictionless banked curve, only two forces act on a car: gravity (down) and the normal force (perpendicular to the road, tilted toward the centre by the bank angle θ).
Important: the acceleration is horizontal, toward the centre of the circle, not along the slope. So choose horizontal and vertical axes (not tilted ones as on an inclined plane), and resolve the normal force:
- Vertical (a = 0): FN cos θ = mg
- Horizontal (toward the centre): FN sin θ = mv² ÷ r
Divide the second equation by the first:
tan θ = v² ÷ (rg), so the design speed is v = √(rg tan θ)
The mass cancels, so every vehicle has the same design speed. Notice also that FN = mg ÷ cos θ, which is more than mg: passengers feel pressed into their seats.
Worked exampleExample 1: Designing a ramp
Highway engineers want a ramp of radius 80 m to be taken safely at 60 km/h with no help from friction. What bank angle is needed?
- v = 60 ÷ 3.6 = 16.7 m/s.
- tan θ = v² ÷ (rg) = 16.7² ÷ (80 × 9.8) = 277.8 ÷ 784 = 0.354.
- θ = tan⁻¹(0.354) = 19.5°.
About 20°. That is steep for a highway: on glare ice, a car stopped on it could slide down to the inside. Real roads combine a smaller bank with friction (see below).
Worked exampleExample 2: The velodrome
A velodrome curve has a radius of 25 m and is banked at 40°. At what speed can a cyclist ride it without relying on sideways friction?
- v = √(rg tan θ) = √(25 × 9.8 × tan 40°) = √(245 × 0.839) = √205.6 = 14.3 m/s.
About 14 m/s (52 km/h). Faster riders lean on friction to stay up; slower riders would slide down the bank, which is why track cyclists must keep their speed up in the turns.
Banked curves with friction
At any speed other than the design speed, friction joins in, acting along the slope:
- Faster than the design speed, the car tends to slide up and out, so friction points down the slope, adding to the inward force.
- Slower, the car tends to slide down and in, so friction points up the slope.
With maximum static friction μs FN included, the same two equations give a range of safe speeds:
v(max)² = rg (tan θ + μs) ÷ (1 − μs tan θ) and v(min)² = rg (tan θ − μs) ÷ (1 + μs tan θ)
If tan θ ≤ μs, the minimum speed is zero: a parked car will not slide down.
Worked exampleExample 3: An icy banked ramp
A ramp of radius 80 m is banked at 15°. On ice, μs = 0.10. Find the design speed and the range of safe speeds.
- Design speed: v = √(80 × 9.8 × tan 15°) = √(784 × 0.268) = 14.5 m/s (52 km/h).
- tan 15° = 0.268. v(max)² = 784 × (0.268 + 0.10) ÷ (1 − 0.0268) = 784 × 0.378 = 296.4, so v(max) = 17.2 m/s (62 km/h).
- v(min)² = 784 × (0.268 − 0.10) ÷ (1 + 0.0268) = 784 × 0.164 = 128.2, so v(min) = 11.3 m/s (41 km/h).
- Since tan 15° = 0.268 > μs = 0.10, a car stopped on this ramp would slide down the bank.
Safe speeds on ice: about 41 km/h to 62 km/h, with an ideal of about 52 km/h. A stopped car would slide toward the inside.
Analysing a technology: banking in highway design (B1.1)
Engineers who design curves must balance several needs:
- Speed: a steeper bank raises the design speed and lets traffic take tighter curves (useful where land is limited, as in a cloverleaf interchange).
- Winter conditions: in snowy regions, roads are banked less steeply, because on ice a slow or stopped vehicle could slide down a steep bank toward the inside of the curve.
- Mixed traffic: trucks with a high centre of mass and slow vehicles must also be safe; drainage of rain and meltwater is improved by some slope.
- Signs: advisory speed signs on ramps suggest a safe speed for the curve in good conditions.
The same physics shows up elsewhere: velodromes and oval race tracks use very steep banks for high speeds; planes and cyclists lean into turns so that the lift or the ground's force tilts toward the centre (tan θ = v² ÷ (rg) again); and some high-speed trains tilt their cars to keep passengers comfortable on curves.
Just for funWhich bank never charges a service fee?
A banked curve: it gives every driver a free inward push, no matter how much they weigh.
Hands-on activityMarbles in a bowl
You need: a large salad bowl or a wok with sloping sides, a marble (keep away from young children), a ruler and a stopwatch.
- Roll the marble around the inside of the bowl so it circles at a steady height near the bottom. Then give it more speed so it circles higher up. Where are the sides steeper?
- Explain using tan θ = v² ÷ (rg): a faster marble needs a steeper bank (or a bigger circle), so it rides higher, where the bowl is steeper.
- For one steady circle, measure the radius (half the width of the bowl at that height) and time 5 laps to find T. Calculate v = 2πr ÷ T and predict the bank angle θ = tan⁻¹(v² ÷ (rg)).
- Estimate the real angle of the bowl's side at that height (hold a ruler against it and use a protractor). How close is your prediction? (The marble rolls, so expect some difference.)
Your turn
1What bank angle lets cars take a curve of radius 50 m at 15 m/s without friction?
tan θ = 15² ÷ (50 × 9.8) = 0.459; θ = 25°.
2Find the design speed of a curve with r = 120 m and θ = 10°, in km/h.
v = √(120 × 9.8 × tan 10°) = 14 m/s, about 52 km/h.
3A car takes a banked curve faster than its design speed. Which way does friction act on it?
Down the slope (toward the inside of the curve), because the car tends to slide up and out.
4Find the normal force on a 1000 kg car going around a 20° banked curve at its design speed. Compare it with the car's weight.
FN = mg ÷ cos 20° = 9800 ÷ 0.940 = 1.04 × 10⁴ N, about 6% more than its weight.
5A plane flying at 70 m/s banks its wings at 25°. Its lift is perpendicular to the wings. Find the radius of its turn.
Like a banked curve: tan θ = v² ÷ (rg), so r = v² ÷ (g tan θ) = 4900 ÷ (9.8 × 0.466) = 1.1 × 10³ m.
6Why is a banked curve safer than a flat one on ice?
On a flat curve, only friction can provide the centripetal force. On a banked curve, the normal force supplies it at the design speed, so little or no friction is needed.
More practice: warm-up
1On a frictionless banked curve, what force supplies the centripetal force?
The horizontal component of the normal force.
2Write the design-speed relationship.
3Does the design speed depend on the vehicle's mass?
4If the design speed must be higher on the same radius, should the bank be steeper or shallower?
5On a banked curve, is the normal force larger or smaller than mg (at the design speed)?
More practice: on your own
1A cyclist rides a velodrome curve of radius 22 m at 15 m/s. What bank angle needs no friction?
tan θ = 225 ÷ (22 × 9.8) = 1.04; θ = 46°.
2A curve of radius 60 m is banked at 12° and μs = 0.50 on dry pavement. Find the maximum safe speed.
tan 12° = 0.213. v² = 60 × 9.8 × (0.213 + 0.50) ÷ (1 − 0.50 × 0.213) = 588 × 0.713 ÷ 0.894 = 469. v = 22 m/s (about 78 km/h).
3A car is stopped on a ramp banked at 15° on glare ice with μs = 0.10. Will it slide? What does this suggest about banking in Ontario?
It slides if tan θ > μs: tan 15° = 0.27 > 0.10, so yes, it slides down toward the inside. Roads in snowy climates are banked less steeply for this reason.
4A cyclist on a flat road takes a turn of radius 20 m at 10 m/s. At what angle from the vertical must she lean?
Like a banked curve: tan θ = v² ÷ (rg) = 100 ÷ 196 = 0.51; θ = 27°.
5What radius of curve, banked at 8°, has a design speed of 100 km/h?
v = 27.8 m/s. r = v² ÷ (g tan θ) = 772 ÷ (9.8 × 0.141) = 5.6 × 10² m.
Challenge
1Derive v(max)² = rg (tan θ + μs) ÷ (1 − μs tan θ) for a banked curve.
At maximum speed, friction μs FN points down the slope. Vertical: FN cos θ − μs FN sin θ = mg. Horizontal: FN sin θ + μs FN cos θ = mv² ÷ r. Divide: v² ÷ (rg) = (sin θ + μs cos θ) ÷ (cos θ − μs sin θ). Divide the top and bottom by cos θ: v² = rg (tan θ + μs) ÷ (1 − μs tan θ).
2For the curve in Example 3 (r = 80 m, θ = 15°), find the range of safe speeds on dry pavement with μs = 0.80.
v(max)² = 784 × (0.268 + 0.80) ÷ (1 − 0.80 × 0.268) = 784 × 1.068 ÷ 0.786 = 1065, so v(max) = 33 m/s (about 117 km/h). For the minimum, tan θ − μs < 0, so v(min) = 0: a stopped car stays put.
3Show that a passenger on a banked curve at the design speed feels pressed into the seat with 1 ÷ cos θ times their weight. Evaluate it for a 40° velodrome bank.
The seat (like the road) pushes perpendicular to the surface with FN = mg ÷ cos θ, which is 1 ÷ cos θ times mg. For 40°: 1 ÷ cos 40° = 1.31, so a cyclist feels about 31% heavier in the turn.
Quiz
1. On a frictionless banked curve at the design speed, the centripetal force is supplied by:
2. The design speed of a banked curve is given by:
3. For banked-curve problems, the axes should be:
4. A car rounds a banked curve more slowly than the design speed. Friction on it points:
5. Why are highway curves in snowy regions banked less steeply?
6. The design speed of a curve is 50 km/h. A truck twice as heavy as a car should take it at:
Big idea: On a banked curve, the normal force leans toward the centre and supplies the centripetal force (B2.6). At other speeds, static friction on the inclined surface helps or hinders (B3.2). Analysing highway banking is an example of analysing a technology that applies circular-motion principles (B1.1).
How to help: On your next drive along a highway ramp (as a passenger), look for the tilt and the advisory speed sign. Ask your teen to explain why the ramp is tilted and why it is not tilted more.
Common difficulties:
- Tilting the axes along the slope, as on an inclined plane, and getting tangled equations. The acceleration here is horizontal.
- Writing FN = mg cos θ (true on a ramp where nothing moves in a circle, but not here: FN = mg ÷ cos θ).
- Getting the friction direction backward for speeds above or below the design speed.
Safety: keep marbles away from young children.