Motion on banked surfaces

Smooth bank

On a surface banked at θ to the horizontal, moving in a horizontal circle of radius r at speed v:

• Vertically: R cos θ = mg.
• Horizontally: R sin θ = mv²/r.

Divide: tan θ = v²/(rg). At this speed no friction is needed, so there is no tendency to slip.

Rough bank

Resolve vertically and horizontally, not along the slope, because the acceleration is horizontal.

• Faster than the frictionless speed: the car tends to slide up the slope, so friction acts down it. At the greatest speed, F = μR: R cos θ - F sin θ = mg and R sin θ + F cos θ = mv²/r.
• Slower: friction acts up the slope. At the least speed, R cos θ + F sin θ = mg and R sin θ - F cos θ = mv²/r.

Results

These give v_max² = rg(sin θ + μ cos θ)/(cos θ - μ sin θ) and v_min² = rg(sin θ - μ cos θ)/(cos θ + μ sin θ). Derive them in the exam rather than quoting.

Worked example

A car moves on a rough track banked at 20° in a horizontal circle of radius 50 m. The coefficient of friction is 0.3. Find the greatest speed without slipping.

1. Friction acts down the slope with F = 0.3R.
2. Vertically: R(cos 20° - 0.3 sin 20°) = mg.
3. Horizontally: R(sin 20° + 0.3 cos 20°) = mv²/50.
4. Divide: v² = 50g(sin 20° + 0.3 cos 20°)/(cos 20° - 0.3 sin 20°).

Answer: v = 19.1 m/s

Check yourself

1. A road is banked at 15° on a bend of radius 80 m. Find the speed at which a car has no tendency to slip, to 3 significant figures.
   Show answer

   14.5 m/s. tan 15° = v²/(80g), v = 14.5 m/s.

2. A track is to be banked so that a car moving at 20 m/s round a bend of radius 100 m needs no friction. Find the angle of banking, to 1 decimal place.
   Show answer

   22.2 °. tan θ = 400/980, θ = 22.2°.

3. In the worked example, find the greatest speed, to 3 significant figures.
   Show answer

   19.1 m/s. v = 19.1 m/s.

4. For the same track, find the least speed at which the car can go round without slipping down the slope, to 3 significant figures.
   Show answer

   5.32 m/s. Friction up the slope: v² = 50g(sin 20° - 0.3 cos 20°)/(cos 20° + 0.3 sin 20°), v = 5.32 m/s.

5. A car of mass 1200 kg moves on a smooth road banked at 15° at the speed for which no friction is needed. Find the normal reaction, to 3 significant figures.
   Show answer

   12200 N. R cos 15° = mg, R = 12200 N.

6. A car goes round a rough banked bend faster than the speed at which no friction is needed. Which way does friction act?
   • Up the slope
   • Along the direction of motion
   • Down the slope
   • There is no friction
   Show answer

   Down the slope. The car tends to slide outwards and up, so friction opposes this by acting down the slope.

7. Why should you resolve vertically and horizontally on a banked track, rather than along and perpendicular to the slope?
   • The acceleration is horizontal, towards the centre, so the vertical equation has no acceleration term
   • The normal reaction is vertical
   • Friction is horizontal
   • It gives a different answer
   Show answer

   The acceleration is horizontal, towards the centre, so the vertical equation has no acceleration term. Resolving vertically gives a balanced equation, and horizontally gives the equation with mv²/r.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M3
• Pearson Edexcel IAL Mechanics M3 (WME03) question paper, June 2025: front-cover instruction on g and accuracy

Last updated 8 November 2026. Spotted a mistake? Tell us.