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Banked bends and the conical pendulum

Key idea: On a frictionless banked bend, the normal force's vertical component balances weight and its horizontal component supplies the inward force. A conical pendulum uses tension in the same way.

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  • GCE A-Level H2 Physics 2027
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On a level road, only friction can push a car sideways round a bend. Racing tracks and railway curves are tilted, or banked, so that the normal force from the road also pushes the car towards the centre. This lesson resolves that tilted normal force to find the speed at which a car can take a banked bend with no friction at all, and applies the same method to a conical pendulum.

Forces on a banked bend

Model a car going round a banked bend at constant speed and constant height, with no friction and no air resistance. Only two real forces act on it: its weight mg, straight down, and the normal force N, at right angles to the road surface.

The road is tilted at angle θ to the horizontal, so the normal force is tilted at θ to the vertical. The centre of the circle is horizontally inwards, so resolve N vertically and horizontally:

  • vertically, the car does not accelerate, so N cos θ = mg;
  • horizontally, the inward component provides the centripetal force, so N sin θ = mv²/r.

Dividing the second equation by the first removes both N and m:

tan θ = v²/rg so v = square root of (rg tan θ)

For a given radius and angle of banking, there is just one speed at which the car needs no friction. At any other speed, friction from the road must act up or down the slope to make up the difference.

Force components on a frictionless banked bendA cross-section of the bend slopes upward towards the outside on the right. The car is represented by a point. Its normal contact force points up and inward, perpendicular to the road, and weight points straight down from the same point. Dashed arrows form a component triangle for the normal: N cos theta vertically upward and N sin theta horizontally inward. Theta is measured from vertical to normal.N sin θN cos θNθmgcar (point model)outsidetowards centre
Force components on a frictionless banked bendA cross-section of the bend slopes upward towards the outside on the right. The car is represented by a point. Its normal contact force points up and inward, perpendicular to the road, and weight points straight down from the same point. Dashed arrows form a component triangle for the normal: N cos theta vertically upward and N sin theta horizontally inward. Theta is measured from vertical to normal.N sin θN cos θNθmgcar (point model)outsidetowards centre
The real forces are N and mg. Dashed arrows resolve N into components; they are not additional interactions. The centre is horizontally inward. Geometry is schematic and force lengths are not a measured scale.

Check your understanding 1 Which component?

On the frictionless bend shown above, which component of the normal force provides the inward resultant? The bank angle is measured to the horizontal.

Inward component
Show answer

N sin θ = mv²/r. N cos θ = mg because there is no vertical acceleration. The components describe the same normal force in two directions; do not draw them as additional interactions.

Worked example 1

Speed and normal force on a banked bend

Problem

A bend of radius 65.0 m is banked at 14.0°. Find the speed at which a car can take the bend with no friction, and the normal force on a car of mass 1.20 × 10³ kg. Take g = 9.81 m s⁻².

Show full solution

  1. Resolve the normal force

    Method

    Write the vertical and horizontal equations.

    Reason

    The car has no vertical acceleration, but it does accelerate towards the centre.

    Working

    Answer

    N cos θ = mg and N sin θ = mv²/r

  2. Speed

    Reason

    Dividing the equations removes N and m.

    Working

    Answer

    v = square root of (rg tan θ) = square root of (65.0(9.81) tan 14.0°) = 12.6 m s⁻¹

  3. Normal force

    Working

    Answer

    N = mg/(cos θ) = 1200(9.81)/(cos 14.0°) = 1.21 × 10⁴ N

Guided practice 2

Speed on a frictionless bend

About 3 min

Problem

A bend of radius 40.0 m is banked at 20.0°. At what speed can a car take it with no friction? Take g = 9.81 m s⁻².

Find the speed

Unit: m s⁻¹

Hints

Hint 1

two equations

Start from N cos θ = mg and N sin θ = mv²/r, then eliminate N.

Show full solution

  1. Speed

    Working

    Answer

    v = square root of (40.0(9.81) tan 20.0°) = 12.0 m s⁻¹. The radius is that of the car’s horizontal circle, not a length measured along the sloping road.

Check your understanding 2 Designing a bend

A banked bend of radius 75.0 m is designed so that cars at 15.0 m s⁻¹ need no friction. Find the angle of banking. How would doubling the mass of the car change the design speed and the normal force? Take g = 9.81 m s⁻².

Show the steps Optional working
Unit: °
Doubling the car mass
Show answer

tan θ = v²/rg = 15.0²/75.0(9.81) = 0.306, so θ = 17.0°. The design speed does not change, because the mass cancels. The normal force doubles, because N = mg/cos θ.

The conical pendulum

A mass on a string that moves round a horizontal circle, with the string sweeping out a cone, is a conical pendulum. The force analysis is the same as for the banked bend, with the tension FT in the string taking the place of the normal force.

Try it yourself 3

The conical pendulum

Minimal support

Problem

A small mass on a light string of length 1.00 m moves steadily round a horizontal circle, with the string at 30.0° to the vertical. Find the radius of the circle and its speed. Take g = 9.81 m s⁻².

Find the radius and speed

Show the steps Optional working
Unit: m
Unit: m s⁻¹

Hints

Hint 1

same structure as the bend

The vertical component of the tension balances the weight; its horizontal component provides the centripetal force.

Show solution step by step

  1. Resolve the tension

    Reason

    The mass has no vertical acceleration, but accelerates horizontally towards the centre.

    Working

    Answer

    First, r = L sin θ = 1.00 sin 30.0° = 0.500 m. Then FT cos θ = mg and FT sin θ = mv²/r, so tan θ = v²/rg

  2. Speed

    Working

    Answer

    v = square root of (rg tan θ) = square root of (0.50(9.81) tan 30.0°) = 1.68 m s⁻¹

Common mistakes

  • Setting N = mg on a banked bend; only the vertical component, N cos θ, balances the weight.
  • Setting the whole normal force or tension equal to mv²/r.
  • Adding N cos θ and N sin θ as extra forces alongside N.
  • Using the string length, or the length along the slope, as the radius of the circle.

Before you move on

Check your understanding 3

A car takes a frictionless banked bend at its design speed. The road’s normal force is 1.25 times the car’s weight. Find the banking angle and the ratio of the inward resultant force to the weight. Does the normal force alone equal the inward resultant?

Unit: °
Show the steps Optional working
Unit: °
Is N alone the inward resultant?
Show answer

Vertical balance gives 1.25mg cos θ = mg, so cos θ = 0.80 and θ = 36.9°. Hence sin θ = 0.60 and the inward resultant is N sin θ = 1.25mg(0.60) = 0.75mg. The ratio is 0.75. Only the horizontal component of the normal force is the inward resultant; its vertical component balances the weight.

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