How to Calculate the Design Speed of a Frictionless Banked Road

A road engineer knows the radius of a bend and how steeply the road is banked. They need to calculate the speed at which a car can travel around the bend without relying on sideways tyre friction.

The important question is not simply how much force the road exerts on the car. It’s how much of that force actually points towards the centre of the circular path.

Before calculating, predict which force supplies the inward acceleration: the car’s weight, the entire normal force, or a component of the normal force?

It’s the inward horizontal component of the normal force. Using this gives the frictionless design-speed equation:

\[ \boxed{v=\sqrt{rg\tan\theta}} \]

where:

• \(v\) is the design speed in \(\text{m s}^{-1}\)
• \(r\) is the radius of the circular path in metres
• \(g\) is gravitational acceleration, taken as \(9.8\ \text{m s}^{-2}\)
• \(\theta\) is the banking angle measured above the horizontal.

The calculation is fairly short. Getting the force components right is what makes it work.

01Identify the real radial force

Imagine a car travelling around a circular road. The outside edge of the road is raised, so the surface slopes upwards as you move away from the centre of the bend.

The road exerts a normal force \(N\) on the car. A normal force always acts perpendicular to the contact surface.

Because the road is tilted, the normal force is tilted too.

There are only two forces in our ideal frictionless model:

• Normal force, \(N\), acting perpendicular to the road.
• Weight, \(mg\), acting vertically downwards.

Friction is absent, but the road still exerts a normal force. Frictionless does not mean force-free.

Which part of the normal force provides centripetal force?

Choose two directions:

• Vertical: positive upwards.
• Radial: positive horizontally towards the centre of the circular path.

The normal force can be resolved into two components:

\[ N_{\text{vertical}}=N\cos\theta \]

\[ N_{\text{radial}}=N\sin\theta \]

The vertical component supports the car against gravity.

The horizontal component points towards the centre of the bend. This is the actual force responsible for the car’s centripetal acceleration.

This distinction is important.

Centripetal force is not an additional force that you draw on a free-body diagram. It describes the net inward force required for circular motion.

For this frictionless banked road, the net inward force is simply \(N\sin\theta\).

Weight has no horizontal radial component because it acts vertically.

02Derive the frictionless design-speed equation

We can now apply Newton’s second law in each direction.

Vertical direction

The car travels around a horizontal circle, so it has no vertical acceleration.

Therefore, the upward and downward forces balance:

\[ \sum F_y=0 \]

\[ N\cos\theta-mg=0 \]

Rearranging:

\[ \boxed{N\cos\theta=mg} \]

Radial direction

The car requires an inward centripetal acceleration:

\[ a_c=\frac{v^2}{r} \]

Newton’s second law gives:

\[ \sum F_r=\frac{mv^2}{r} \]

Since the inward component of the normal force supplies this acceleration:

\[ \boxed{N\sin\theta=\frac{mv^2}{r}} \]

We now have two equations:

\[ N\cos\theta=mg \]

\[ N\sin\theta=\frac{mv^2}{r} \]

Divide the radial equation by the vertical equation:

\[ \frac{N\sin\theta}{N\cos\theta} = \frac{mv^2/r}{mg} \]

Simplifying:

\[ \tan\theta=\frac{v^2}{rg} \]

Rearrange to make \(v\) the subject:

\[ v^2=rg\tan\theta \]

Therefore:

\[ \boxed{v=\sqrt{rg\tan\theta}} \]

This is the equation you need to calculate the design speed of a frictionless banked road.

Notice that the mass has cancelled.

A heavier car needs a greater inward force, but it also experiences a proportionally greater normal force. Both effects scale with mass, so the ideal design speed is independent of the vehicle’s mass.

03A reliable method for calculating design speed

When you encounter a banked-road question in an HSC Physics exam, follow this process.

There are two details worth checking before substituting.

First: Make sure you have the correct angle.

The banking angle is the angle between the road surface and the horizontal.

The normal force makes the same angle with the vertical. These are equal because the normal force is perpendicular to the road, and the vertical is perpendicular to the horizontal.

If a diagram gives the angle between the normal force and the horizontal, you may need to subtract it from \(90^\circ\).

Second: Make sure you have the correct radius.

Use the radius of the circular path followed by the vehicle, not automatically the radius of the inner edge of the road.

If the question gives the radius of the vehicle’s path or the road’s centreline, use that value.

When the banking angle comes from road measurements

Sometimes a question gives the height difference between the two edges of a road rather than the banking angle itself.

If the outside edge is raised by a vertical height \(h\) across a horizontal distance \(w\), use:

\[ \tan\theta=\frac{h}{w} \]

You can then substitute directly into the design-speed equation:

\[ v=\sqrt{rg\left(\frac{h}{w}\right)} \]

This is particularly useful when a question provides a road cross-section or survey measurements.

Make sure \(w\) is the horizontal distance. If the question instead gives the sloping road width, the trigonometric relationship is different.

04The common mistake: Using the entire normal force

A student might write:

\[ N=\frac{mv^2}{r} \]

This looks reasonable because the normal force is the only contact force acting on the frictionless car.

But it is incorrect.

The normal force is tilted. Only its horizontal component points towards the centre of the circular path.

The correct equation is:

\[ N\sin\theta=\frac{mv^2}{r} \]

The remaining component, \(N\cos\theta\), balances the car’s weight.

This comparison also explains why a flat frictionless road cannot provide the inward force needed for a non-zero circular speed.

On a flat road, the normal force points vertically upwards. It has no horizontal component.

On a banked road, tilting the surface redirects the normal force, allowing it to provide both vertical support and horizontal acceleration.

A useful boundary check is to put \(\theta=0^\circ\) into the design-speed equation:

\[ v=\sqrt{rg\tan0^\circ}=0 \]

That is exactly what we expect from a flat, frictionless road: it cannot support a non-zero circular speed.

05Worked examples

06Questions and solutions

Question 1

A circular road has a radius of \(85.0\ \text{m}\) and a banking angle of \(11.0^\circ\).

Assuming the road is frictionless, calculate the design speed of a car travelling around the bend.

Give your answer in:

a. \(\text{m s}^{-1}\)

b. \(\text{km/h}\)

Solution 1

The design speed is \(12.7\ \text{m s}^{-1}\), or approximately \(45.8\ \text{km/h}\).

For a frictionless banked road:

\[ v=\sqrt{rg\tan\theta} \]

Substitute:

\[ v=\sqrt{(85.0)(9.8)\tan11.0^\circ} \]

\[ v=\sqrt{161.92} \]

\[ \boxed{v=12.7\ \text{m s}^{-1}} \]

Convert to kilometres per hour:

\[ v=12.7\times3.6 \]

\[ \boxed{v=45.8\ \text{km/h}} \]

The important point is that the radius and banking angle determine the design speed. The vehicle’s mass is not required.

Question 2

A proposed banked road has a radius of \(150\ \text{m}\).

A survey establishes that the outside edge is \(0.750\ \text{m}\) higher than the inside edge, across a horizontal distance of \(6.00\ \text{m}\).

A car travels around the bend at \(15.0\ \text{m s}^{-1}\).

Assuming there is no friction, determine:

a. the banking angle,

b. the frictionless design speed, and

c. whether the car can maintain the specified horizontal circular path at \(15.0\ \text{m s}^{-1}\).

Solution 2

The banking angle is \(7.13^\circ\), and the frictionless design speed is \(13.6\ \text{m s}^{-1}\). The car travelling at \(15.0\ \text{m s}^{-1}\) cannot maintain the specified horizontal circular path without an additional inward force.

Part a: Calculate the banking angle.

Use the survey measurements:

\[ \tan\theta=\frac{h}{w} \]

\[ \tan\theta=\frac{0.750}{6.00} \]

\[ \tan\theta=0.125 \]

Therefore:

\[ \theta=\tan^{-1}(0.125) \]

\[ \boxed{\theta=7.13^\circ} \]

Part b: Calculate the design speed.

Use:

\[ v=\sqrt{rg\tan\theta} \]

Substitute:

\[ v=\sqrt{(150)(9.8)(0.125)} \]

\[ v=\sqrt{183.75} \]

\[ \boxed{v=13.6\ \text{m s}^{-1}} \]

Part c: Check whether the car can maintain the path.

The actual speed is:

\[ v_{\text{actual}}=15.0\ \text{m s}^{-1} \]

The design speed is:

\[ v_{\text{design}}=13.6\ \text{m s}^{-1} \]

These are not equal.

We can check the mismatch using radial acceleration.

At the actual speed, the required inward acceleration is:

\[ a_{\text{required}}=\frac{v^2}{r} \]

\[ a_{\text{required}}=\frac{15.0^2}{150} \]

\[ a_{\text{required}}=1.50\ \text{m s}^{-2} \]

However, the frictionless bank provides an inward acceleration of:

\[ a_{\text{bank}}=g\tan\theta \]

\[ a_{\text{bank}}=(9.8)(0.125) \]

\[ a_{\text{bank}}=1.225\ \text{m s}^{-2} \]

The required acceleration is greater than the available inward acceleration.

Therefore, without friction, the car cannot maintain the specified circular path at this speed.

This is an important distinction: knowing the design speed lets you determine whether a particular speed is compatible with the frictionless model. It does not mean the car can automatically travel at any speed on that road.

Question 3

Two vehicles travel around the same frictionless banked curve.

The road has:

• Radius \(70.0\ \text{m}\)
• Banking angle \(14.0^\circ\)

Vehicle A has a mass of \(900\ \text{kg}\), while Vehicle B has a mass of \(1800\ \text{kg}\).

Determine:

a. the design speed for both vehicles,

b. the centripetal acceleration of each vehicle, and

c. the inward radial force experienced by each vehicle.

Explain why the two vehicles have the same design speed despite having different masses.

Solution 3

Both vehicles have a design speed of \(13.1\ \text{m s}^{-1}\) and a centripetal acceleration of \(2.44\ \text{m s}^{-2}\). Vehicle A experiences an inward force of approximately \(2.20\ \text{kN}\), while Vehicle B experiences \(4.40\ \text{kN}\).

Part a: Calculate the design speed.

Use:

\[ v=\sqrt{rg\tan\theta} \]

Substitute:

\[ v=\sqrt{(70.0)(9.8)\tan14.0^\circ} \]

\[ v=\sqrt{171.04} \]

\[ \boxed{v=13.1\ \text{m s}^{-1}} \]

The equation does not contain mass, so both vehicles have the same design speed.

Part b: Calculate the centripetal acceleration.

Use:

\[ a_c=\frac{v^2}{r} \]

Substitute:

\[ a_c=\frac{(13.1)^2}{70.0} \]

Using the unrounded speed:

\[ \boxed{a_c=2.44\ \text{m s}^{-2}} \]

Both vehicles have the same speed and radius, so their centripetal accelerations are equal.

Part c: Calculate the radial force.

Newton’s second law gives:

\[ F_r=ma_c \]

For Vehicle A:

\[ F_A=(900)(2.44) \]

\[ \boxed{F_A=2.20\times10^3\ \text{N}} \]

For Vehicle B:

\[ F_B=(1800)(2.44) \]

\[ \boxed{F_B=4.40\times10^3\ \text{N}} \]

Vehicle B requires twice the inward force because its mass is twice as large.

We can also verify that these forces are supplied by the normal force components.

From vertical equilibrium:

\[ N\cos\theta=mg \]

Therefore:

\[ N=\frac{mg}{\cos\theta} \]

For Vehicle A:

\[ N_A=\frac{(900)(9.8)}{\cos14^\circ} \]

\[ N_A=9.09\times10^3\ \text{N} \]

Its inward component is:

\[ N_A\sin14^\circ\approx2.20\times10^3\ \text{N} \]

For Vehicle B:

\[ N_B=\frac{(1800)(9.8)}{\cos14^\circ} \]

\[ N_B=1.82\times10^4\ \text{N} \]

Its inward component is:

\[ N_B\sin14^\circ\approx4.40\times10^3\ \text{N} \]

This confirms the central idea: the heavier vehicle needs a larger normal force and a larger radial force, but the ratio between the two remains unchanged.

That is why its frictionless design speed is the same.

Question 4

A physics student receives a diagram of a frictionless banked road.

The normal force acting on the vehicle is directed \(78.0^\circ\) above the horizontal, towards the centre of the bend.

The road has a circular path radius of \(95.0\ \text{m}\).

Calculate the design speed.

Explain why substituting \(78.0^\circ\) directly into the design-speed equation would be incorrect.

Solution 4

The design speed is \(14.1\ \text{m s}^{-1}\), or approximately \(50.6\ \text{km/h}\).

The correct banking angle is \(12.0^\circ\), not \(78.0^\circ\).

The design-speed equation requires the angle between the road surface and the horizontal.

The question instead gives the angle between the normal force and the horizontal.

Because the normal force is perpendicular to the road, these two angles are complementary.

Step 1: Calculate the banking angle.

\[ \theta=90^\circ-78.0^\circ \]

\[ \boxed{\theta=12.0^\circ} \]

Step 2: Calculate the design speed.

Use:

\[ v=\sqrt{rg\tan\theta} \]

Substitute:

\[ v=\sqrt{(95.0)(9.8)\tan12.0^\circ} \]

\[ v=\sqrt{197.89} \]

\[ \boxed{v=14.1\ \text{m s}^{-1}} \]

Convert to kilometres per hour:

\[ v=14.1\times3.6 \]

\[ \boxed{v=50.6\ \text{km/h}} \]

The tempting mistake is to use \(78.0^\circ\) as the banking angle.

That would give:

\[ v=\sqrt{(95.0)(9.8)\tan78.0^\circ} \]

\[ v\approx66.2\ \text{m s}^{-1} \]

This is incorrect because the equation uses the road’s angle above the horizontal, not the normal force’s angle above the horizontal.

Always identify what the angle is measured between before substituting it.

Question 5

A steeply banked closed test track has a circular section with:

• Radius \(80.0\ \text{m}\)
• Banking angle \(30.0^\circ\)

A car travels around this section at its frictionless design speed.

The track then enters a transition section where the local radius increases to \(240\ \text{m}\).

Assume:

• The car’s speed remains constant.
• The road is frictionless.
• Each section can be treated as a local horizontal circular path.
• The banking angle can be adjusted independently along the transition.

An engineer proposes reducing the banking angle from \(30.0^\circ\) to \(10.0^\circ\), reasoning that the radius has tripled, so the banking angle should be divided by three.

Investigate this proposal.

a. Calculate the car’s initial speed.

b. Determine the exact banking angle required at the larger radius for the car to maintain the same speed without friction.

c. Calculate the frictionless design speed of the proposed \(10.0^\circ\) bank.

d. Determine whether the proposed design allows the car to maintain the prescribed circular path. Explain the physical reason.

Solution 5

The initial speed is \(21.3\ \text{m s}^{-1}\). The correct final banking angle is approximately \(10.9^\circ\), not \(10.0^\circ\). The proposed design therefore cannot maintain the prescribed frictionless circular path at the car’s unchanged speed.

This problem tests whether you understand the relationship between banking angle, radius, and speed, rather than simply scaling the numbers.

Part a: Calculate the initial speed.

The car initially travels at the frictionless design speed for:

\[ r_1=80.0\ \text{m} \]

\[ \theta_1=30.0^\circ \]

Use:

\[ v_1=\sqrt{r_1g\tan\theta_1} \]

Substitute:

\[ v_1=\sqrt{(80.0)(9.8)\tan30.0^\circ} \]

\[ v_1=\sqrt{452.44} \]

\[ \boxed{v_1=21.3\ \text{m s}^{-1}} \]

The question states that the speed remains constant, so the car must still travel at approximately \(21.3\ \text{m s}^{-1}\) in the larger-radius section.

Part b: Calculate the required final banking angle.

The second section has:

\[ r_2=240\ \text{m} \]

For the car to maintain its speed without friction:

\[ \tan\theta_2=\frac{v_1^2}{r_2g} \]

Substitute the original design-speed relationship:

\[ \tan\theta_2 = \frac{r_1g\tan\theta_1}{r_2g} \]

The gravitational acceleration cancels:

\[ \tan\theta_2 = \frac{r_1}{r_2}\tan\theta_1 \]

Substitute:

\[ \tan\theta_2 = \frac{80.0}{240}\tan30.0^\circ \]

\[ \tan\theta_2=0.19245 \]

Therefore:

\[ \theta_2=\tan^{-1}(0.19245) \]

\[ \boxed{\theta_2=10.9^\circ} \]

The correct banking angle is approximately \(10.9^\circ\).

Part c: Calculate the design speed for the proposed bank.

The engineer proposes a banking angle of \(10.0^\circ\).

The design speed for this angle and the larger radius is:

\[ v_{\text{design},2} = \sqrt{r_2g\tan10.0^\circ} \]

Substitute:

\[ v_{\text{design},2} = \sqrt{(240)(9.8)\tan10.0^\circ} \]

\[ v_{\text{design},2} = \sqrt{414.7} \]

\[ \boxed{v_{\text{design},2}=20.4\ \text{m s}^{-1}} \]

Part d: Decide whether the proposed design works.

The car’s actual speed is:

\[ v_{\text{actual}}=21.3\ \text{m s}^{-1} \]

The proposed design speed is:

\[ v_{\text{design},2}=20.4\ \text{m s}^{-1} \]

These are different.

We can verify the mismatch by comparing the required inward acceleration with the acceleration provided by the bank.

At the larger radius, the car requires:

\[ a_{\text{required}}=\frac{v^2}{r_2} \]

\[ a_{\text{required}}=\frac{452.44}{240} \]

\[ a_{\text{required}}=1.885\ \text{m s}^{-2} \]

The proposed bank provides:

\[ a_{\text{bank}}=g\tan10.0^\circ \]

\[ a_{\text{bank}}=(9.8)\tan10.0^\circ \]

\[ a_{\text{bank}}=1.728\ \text{m s}^{-2} \]

The required inward acceleration is greater than the acceleration supplied by the normal force.

There is therefore an inward force deficit of approximately:

\[ 1.885-1.728=0.157\ \text{m s}^{-2} \]

Without friction, the car cannot maintain the prescribed horizontal circular path.

The subtle mistake in the engineer’s reasoning is assuming that banking angle is directly proportional to the inverse of radius.

The exact relationship is:

\[ \tan\theta=\frac{v^2}{rg} \]

It is the tangent of the banking angle that changes in proportion to the inverse of radius, not the angle itself.

For small angles measured in radians, \(\tan\theta\approx\theta\), which can make direct angle scaling a reasonable approximation. But it is not exact, particularly for a relatively steep bank.

This is why the original \(30.0^\circ\) bank must be recalculated using trigonometry rather than simply divided by three.

06Where to go next

Calculating the frictionless design speed gives you one precise result: the speed at which the normal force alone can provide the required centripetal force.

Real roads are different. Tyres provide static friction, allowing vehicles to travel at speeds above or below the ideal design speed, provided sufficient grip is available.

The next step is to understand how friction changes the force diagram, why its direction depends on the vehicle’s speed, and how to calculate the minimum and maximum speeds a banked road can safely accommodate.

For that broader treatment, see Banked Tracks: Normal Force and Design Speed in HSC Physics.

Syllabus context: This explanation follows the 2017 Physics Stage 6 syllabus treatment of Circular Motion. The NSW Physics 11-12 Syllabus (2025) begins implementation in 2027, so students should check which syllabus version their school is teaching. See the NSW Physics syllabus information.