Why Steeper Banked Tracks Support Higher Design Speeds

Two bends have the same radius of \(100\ \text{m}\). One is banked at \(8^\circ\), while the other is banked at \(20^\circ\). Which bend allows a car to travel faster without relying on sideways tyre friction?

The \(20^\circ\) bend. Its steeper surface tilts the normal force further towards the centre of the circular path, increasing the inward force available to turn the car.

For the same radius, a steeper bank therefore supports a higher frictionless design speed:

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

But there is an important qualification: this comparison assumes the radius stays constant. If both the radius and banking angle change, you need to consider both.

01Which force actually turns the car?

Before looking at why the angle matters, consider what happens to the forces on a car travelling around a banked bend.

The outside edge of the road is raised, so the road surface slopes upwards as you move away from the centre of the turn.

The road pushes against the car with a contact force called the normal force, \(N\). A normal force always acts perpendicular to the contact surface.

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

Which part of this force actually points towards the centre of the bend: the entire normal force or just one component?

It is the horizontal, inward component.

Let’s establish the directions:

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

The banking angle, \(\theta\), is measured between the road surface and the horizontal.

The normal force makes the same angle with the vertical.

We can therefore resolve it into two components:

• \(N\cos\theta\): the upward component.
• \(N\sin\theta\): the inward radial component.

The car’s weight, \(mg\), acts vertically downwards. It has no horizontal radial component.

This gives us an important distinction:

In the frictionless model, the actual inward radial force is \(N\sin\theta\).

Centripetal force is not an additional force acting on the car. It is the name given to the net inward force required for circular motion.

For a frictionless banked track, the normal force supplies that inward force.

02Why increasing the banking angle increases design speed

Let’s connect the force components to the car’s speed.

Assume:

• The car follows a horizontal circular path of constant radius \(r\).
• There is no sideways friction.
• The car has no vertical acceleration.
• Gravitational acceleration is \(g=9.8\ \text{m s}^{-2}\).

In the vertical direction, the forces balance:

\[ N\cos\theta=mg \]

In the radial direction, the inward normal force component provides the required centripetal force:

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

Here:

• \(N\) is the normal force in newtons.
• \(m\) is the car’s mass in kilograms.
• \(v\) is its speed in metres per second.
• \(r\) is the radius of its circular path in metres.
• \(\theta\) is the banking angle.

Divide the radial equation by the vertical equation:

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

Simplifying gives:

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

Rearranging:

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

This is the frictionless design-speed equation.

Now we can see exactly why a steeper bank supports a higher design speed.

For a fixed radius and gravitational acceleration:

\[ v^2\propto\tan\theta \]

As the banking angle increases, \(\tan\theta\) increases. The design speed must therefore increase too.

There is another useful way to understand this.

From the vertical force equation:

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

Substituting into the radial force equation gives:

\[ F_r=mg\tan\theta \]

where \(F_r\) is the inward radial force.

A steeper bank increases the inward force available relative to the car’s weight. The car can therefore travel faster while following the same circular path.

Notice that the mass has cancelled. In this ideal model, a heavier car has the same frictionless design speed as a lighter car.

How much does the design speed actually change?

Let’s use an illustrative comparison with a constant radius of \(100\ \text{m}\).

The following values come from the frictionless design-speed equation, using \(g=9.8\ \text{m s}^{-2}\).

The relationship is not linear.

For example, doubling the banking angle from \(10^\circ\) to \(20^\circ\) does not double the design speed.

Instead:

\[ \frac{v_{20}}{v_{10}} = \sqrt{\frac{\tan20^\circ}{\tan10^\circ}} \]

\[ \frac{v_{20}}{v_{10}}\approx1.44 \]

The design speed increases by approximately \(44\%\).

The reason is that speed depends on the square root of the tangent of the banking angle, not directly on the angle itself.

All values in the table are illustrative theoretical predictions, not recommended real-world road speeds.

03What if the radius also changes?

This is where students can make a convincing mistake.

A student might look at two bends, notice that one is more steeply banked, and immediately conclude that it must have the higher design speed.

That conclusion is only reliable if the radii are equal.

Remember:

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

Both the radius and the banking angle affect the design speed.

For two different curves, the ratio is:

\[ \frac{v_2}{v_1} = \sqrt{ \frac{r_2\tan\theta_2} {r_1\tan\theta_1} } \]

This gives us a useful decision rule.

The key is to avoid judging a track by its steepness alone.

04A model experiment: Testing the effect of banking angle

How could we test whether a steeper bank really produces a higher frictionless design speed?

We can build a small-scale model of a banked circular track.

Setting up the apparatus

Start with an adjustable circular track.

The track should have:

• A clearly marked centre.
• A circular path with a measurable radius.
• An outside edge that can be raised to change the banking angle.
• A small trolley or low-friction cart.

Measure the radius from the centre of the circular path to the centreline of the trolley’s path.

Use an inclinometer to measure the banking angle relative to the horizontal.

An overhead camera can record the trolley’s motion. Its lap time can be used to calculate speed.

For one complete revolution, the distance travelled is the circumference:

\[ d=2\pi r \]

If the trolley takes time \(T\) to complete one revolution:

\[ v=\frac{2\pi r}{T} \]

What should happen?

Keep the radius constant and increase the banking angle.

The prediction from our force model is that the speed at which the trolley requires no sideways contact force will increase.

For example, consider a model track with:

\[ r=1.00\ \text{m} \]

At a banking angle of \(10^\circ\):

\[ v=\sqrt{(1.00)(9.8)\tan10^\circ} \]

\[ v=1.31\ \text{m s}^{-1} \]

The predicted time for one revolution is:

\[ T=\frac{2\pi(1.00)}{1.31} \]

\[ T=4.78\ \text{s} \]

Now increase the banking angle to \(20^\circ\), keeping the radius unchanged:

\[ v=\sqrt{(1.00)(9.8)\tan20^\circ} \]

\[ v=1.89\ \text{m s}^{-1} \]

The predicted lap time becomes:

\[ T=\frac{2\pi(1.00)}{1.89} \]

\[ T=3.33\ \text{s} \]

These are calculated predictions, not recorded experimental measurements.

The shorter lap time at \(20^\circ\) is consistent with the model’s prediction of a higher frictionless design speed.

What could go wrong with the experiment?

Simply timing the trolley is not enough.

A trolley might complete a lap at a different speed while friction supplies some of the required radial force. Its path might also drift towards the inside or outside of the track, changing the actual radius.

A better investigation would:

• Use a low-friction surface to minimise sideways friction.
• Record the trolley’s path from above.
• Measure the actual radius of its path.
• If available, use a force sensor to check when the sideways contact force is approximately zero.
• Repeat the measurements at each banking angle.

This matters because the experiment is testing a specific prediction: the speed at which the track’s normal force alone can provide the required inward force.

Real apparatus will never be perfectly frictionless, so measured results may differ slightly from the theoretical predictions.

05Extension: What happens when the car travels at the wrong speed?

The frictionless design speed is the speed at which no sideways tyre friction is required.

It is not necessarily the only speed at which a real car can travel around the bend.

Tyre friction can help the car follow the circular path at speeds above or below the design speed.

Consider a banked road with its outside edge raised.

Below the design speed

The car tends to slide down the bank, towards the inside of the curve.

Static friction acts up the slope, towards the outside edge.

Its horizontal component points outwards, reducing the net inward radial force.

Above the design speed

The car tends to move up the bank, towards the outside edge.

Static friction acts down the slope, towards the inside of the curve.

Its horizontal component points inwards, increasing the net inward radial force.

Using positive radial direction towards the centre, the radial force equation becomes different in each case.

Below design speed:

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

Above design speed:

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

Here, \(f\) is the magnitude of the sideways friction force.

Notice that friction changes the vertical force balance as well, because it has a vertical component.

This is why the frictionless design-speed equation should not be treated as a maximum safe speed. It describes one particular condition: zero sideways friction.

For a broader explanation of these force directions and the design-speed relationship, see Banked Tracks: Normal Force and Design Speed in HSC Physics.

08Questions and solutions

Question 1

A circular bend has a radius of \(95.0\ \text{m}\) and a banking angle of \(12.0^\circ\).

Assuming there is no sideways friction, calculate the frictionless design speed of a car travelling around the bend.

Give your answer in:

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

b. \(\text{km h}^{-1}\)

Solution 1

The frictionless design speed is \(14.1\ \text{m s}^{-1}\), or approximately \(50.7\ \text{km h}^{-1}\).

The inward component of the normal force must provide the centripetal force.

Step 1: Start with the design-speed equation.

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

Step 2: Substitute the known values.

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

\[ v=\sqrt{197.99} \]

\[ v=14.07\ \text{m s}^{-1} \]

Step 3: Convert to kilometres per hour.

Multiply by \(3.6\):

\[ v=14.07\times3.6 \]

\[ v=50.65\ \text{km h}^{-1} \]

Therefore:

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

or:

\[ v=50.7\ \text{km h}^{-1} \]

At this speed, the horizontal component of the normal force provides exactly the inward force required for the car to follow the circular path without sideways friction.

Question 2

Two proposed circular bends have different radii and banking angles.

Both tracks are assumed to be frictionless.

a. Calculate the frictionless design speed for each track.

b. Determine which track supports the higher design speed.

c. Calculate the percentage by which the higher design speed exceeds the lower one.

d. Explain why comparing the banking angles alone would be insufficient.

Solution 2

Track A supports the higher design speed, at approximately \(16.9\ \text{m s}^{-1}\), compared with \(16.1\ \text{m s}^{-1}\) for Track B. Its design speed is approximately \(4.9\%\) higher.

Although Track A has a smaller radius, its steeper banking angle produces a slightly larger value of \(r\tan\theta\).

Step 1: Calculate the design speed for Track A.

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

Substitute:

\[ v_A=\sqrt{(80.0)(9.8)\tan20.0^\circ} \]

\[ v_A=\sqrt{285.35} \]

\[ v_A=16.9\ \text{m s}^{-1} \]

Step 2: Calculate the design speed for Track B.

\[ v_B=\sqrt{(150)(9.8)\tan10.0^\circ} \]

\[ v_B=\sqrt{259.20} \]

\[ v_B=16.1\ \text{m s}^{-1} \]

Step 3: Compare the speeds.

Track A has the higher design speed.

Calculate the percentage increase relative to Track B:

\[ \%\text{ increase} = \frac{v_A-v_B}{v_B}\times100 \]

\[ = \frac{16.9-16.1}{16.1}\times100 \]

\[ \approx4.9\% \]

Step 4: Explain the effect of radius.

The tempting prediction is that Track A must be much faster because it has a substantially steeper banking angle.

However, Track B has a larger radius, which also increases its design speed.

The correct comparison uses:

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

Track A’s steeper angle compensates for its smaller radius, but only produces a modest speed advantage.

This is why you must consider both quantities when comparing different track designs.

Question 3

A \(1200\ \text{kg}\) car travels at \(16.0\ \text{m s}^{-1}\) around a circular bend of radius \(100\ \text{m}\).

Initially, the road is banked at the angle required for the car to travel without sideways friction.

a. Calculate the required banking angle.

b. Calculate the normal force acting on the car.

c. Calculate the inward radial force acting on the car.

The engineer then increases the banking angle to \(20.0^\circ\), without changing the radius. The car continues travelling at \(16.0\ \text{m s}^{-1}\).

d. Calculate the new frictionless design speed.

e. If the car continues following the horizontal circular path, determine the direction of static friction required. Explain your reasoning.

Use \(g=9.8\ \text{m s}^{-2}\).

Solution 3

The original banking angle is \(14.6^\circ\), the normal force is approximately \(1.22\times10^4\ \text{N}\), and the inward radial force is \(3.07\times10^3\ \text{N}\).

After increasing the bank to \(20.0^\circ\), the car is travelling below the new design speed. Static friction must therefore act up the bank, towards the outside edge.

Step 1: Calculate the original banking angle.

Start with:

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

Substitute:

\[ \tan\theta= \frac{(16.0)^2}{(100)(9.8)} \]

\[ \tan\theta=\frac{256}{980} \]

\[ \tan\theta=0.2612 \]

Take the inverse tangent:

\[ \theta=\tan^{-1}(0.2612) \]

\[ \theta=14.6^\circ \]

Step 2: Calculate the normal force.

In the vertical direction, the normal force balances the weight:

\[ N\cos\theta=mg \]

Rearrange:

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

Substitute:

\[ N=\frac{(1200)(9.8)}{\cos14.6^\circ} \]

\[ N\approx1.22\times10^4\ \text{N} \]

Step 3: Calculate the actual inward radial force.

For the original frictionless design:

\[ F_r=N\sin\theta \]

Alternatively, use Newton’s second law:

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

Substitute:

\[ F_r=\frac{(1200)(16.0)^2}{100} \]

\[ F_r=3072\ \text{N} \]

Therefore:

\[ F_r=3.07\times10^3\ \text{N} \]

This is the horizontal inward component of the normal force in the original frictionless situation.

Step 4: Calculate the new design speed.

The new banking angle is \(20.0^\circ\).

Use:

\[ v_{\text{design}}=\sqrt{rg\tan\theta} \]

Substitute:

\[ v_{\text{design}} = \sqrt{(100)(9.8)\tan20.0^\circ} \]

\[ v_{\text{design}}=\sqrt{356.69} \]

\[ v_{\text{design}}=18.9\ \text{m s}^{-1} \]

The car is travelling at \(16.0\ \text{m s}^{-1}\), which is below the new design speed.

Step 5: Determine the direction of friction.

Below the design speed, the car tends to slide down the bank towards the inside of the curve.

Static friction opposes this tendency.

Therefore, friction acts up the slope, towards the raised outside edge.

Its horizontal component acts outwards, reducing the net inward radial force.

The radial force equation is therefore:

\[ N\sin20^\circ-f\cos20^\circ = \frac{mv^2}{r} \]

The important point is that increasing the banking angle changes the frictionless design speed. A car travelling at its old design speed may now need friction to maintain the same circular path.

Question 4

A road engineer is designing a bend with a radius of \(150\ \text{m}\).

The road has a horizontal width of \(6.0\ \text{m}\), and its outside edge is currently \(0.60\ \text{m}\) higher than its inside edge.

The engineer wants the road to support a frictionless design speed of \(17.0\ \text{m s}^{-1}\).

a. Calculate the current banking angle.

b. Calculate the current frictionless design speed.

c. Determine the banking angle required to achieve the target speed.

d. Calculate the new height difference required between the road edges.

Assume the horizontal road width and radius remain unchanged.

Solution 4

The existing design speed is approximately \(12.1\ \text{m s}^{-1}\). To achieve \(17.0\ \text{m s}^{-1}\), the banking angle must increase to approximately \(11.1^\circ\), requiring a height difference of approximately \(1.18\ \text{m}\).

Step 1: Calculate the current banking angle.

Use the road cross-section:

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

Substitute:

\[ \tan\theta=\frac{0.60}{6.0} \]

\[ \tan\theta=0.100 \]

Therefore:

\[ \theta=\tan^{-1}(0.100) \]

\[ \theta=5.71^\circ \]

Step 2: Calculate the current design speed.

Use:

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

Substitute:

\[ v=\sqrt{(150)(9.8)(0.100)} \]

\[ v=\sqrt{147} \]

\[ v=12.1\ \text{m s}^{-1} \]

Step 3: Find the banking angle needed for the target speed.

Rearrange:

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

Substitute:

\[ \tan\theta= \frac{(17.0)^2}{(150)(9.8)} \]

\[ \tan\theta=\frac{289}{1470} \]

\[ \tan\theta=0.1966 \]

Therefore:

\[ \theta=\tan^{-1}(0.1966) \]

\[ \theta=11.1^\circ \]

Step 4: Calculate the new height difference.

Use:

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

Rearrange:

\[ h=w\tan\theta \]

Substitute:

\[ h=(6.0)(0.1966) \]

\[ h=1.18\ \text{m} \]

The new height difference must be approximately \(1.18\ \text{m}\).

The engineer therefore needs to increase the height difference by:

\[ 1.18-0.60=0.58\ \text{m} \]

The result demonstrates how a banking angle can be calculated from a target speed rather than simply being supplied in a question.

Question 5

An indoor test circuit has two concentric lanes around a circular bend.

The inner lane has:

• A radius of \(70.0\ \text{m}\).
• A banking angle of \(18.0^\circ\).

The outer lane has a radius of \(100\ \text{m}\).

A car travels along the inner lane at its frictionless design speed.

The track designer wants the car to be able to move from the inner lane to the outer lane without changing speed or requiring sideways tyre friction.

Consider two possible designs for the outer lane.

• Design A keeps the outer lane at the same \(18.0^\circ\) banking angle.
• Design B changes the local banking angle of the outer lane.

Assume each lane has a constant local banking angle, the car follows a horizontal circular path, and the transition between lanes can be ignored.

a. Calculate the frictionless design speed of the inner lane.

b. Calculate the frictionless design speed of the outer lane under Design A.

c. If the car moves to the outer lane under Design A without changing speed, determine whether it is travelling above or below the outer lane’s design speed. State the direction of static friction required to maintain the circular path.

d. Calculate the banking angle required for the outer lane under Design B so that the car can maintain its original speed without sideways friction.

e. Explain why the outer lane requires a smaller banking angle, even though it has a larger radius.

Solution 5

The inner lane’s design speed is \(14.9\ \text{m s}^{-1}\). With the same \(18.0^\circ\) banking angle, the outer lane has a higher design speed of \(17.8\ \text{m s}^{-1}\).

At the original speed, the car is below the outer lane’s design speed, so friction must act up the bank.

To allow the car to travel at the same speed without sideways friction, the outer lane must have a banking angle of approximately \(12.8^\circ\).

The surprising result is that the outer lane needs to be less steeply banked, not more steeply banked.

Step 1: Calculate the inner lane’s frictionless design speed.

Use:

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

Substitute:

\[ v_{\text{inner}} = \sqrt{(70.0)(9.8)\tan18.0^\circ} \]

\[ v_{\text{inner}}=\sqrt{222.89} \]

\[ v_{\text{inner}}=14.9\ \text{m s}^{-1} \]

This is the speed the car maintains when it moves towards the outer lane.

Step 2: Calculate the outer lane’s design speed under Design A.

The outer lane has the same banking angle but a larger radius.

\[ v_{\text{outer}} = \sqrt{rg\tan\theta} \]

Substitute:

\[ v_{\text{outer}} = \sqrt{(100)(9.8)\tan18.0^\circ} \]

\[ v_{\text{outer}}=\sqrt{318.42} \]

\[ v_{\text{outer}}=17.8\ \text{m s}^{-1} \]

The outer lane’s design speed is higher than the car’s current speed.

Step 3: Determine the direction of friction.

The car is travelling at:

\[ 14.9\ \text{m s}^{-1} \]

But the outer lane’s design speed is:

\[ 17.8\ \text{m s}^{-1} \]

Therefore, the car is travelling below the outer lane’s design speed.

It tends to slide down the bank, towards the inside of the curve.

Static friction must act up the bank, towards the raised outside edge.

Its horizontal component points outwards, reducing the net inward force to the amount required for the car’s actual speed.

Step 4: Calculate the banking angle required under Design B.

The car must maintain its original speed:

\[ v=14.9\ \text{m s}^{-1} \]

The outer radius is:

\[ r=100\ \text{m} \]

For a frictionless banked track:

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

Substitute the original speed calculated in Step 1:

\[ \tan\theta_{\text{outer}} = \frac{(14.9)^2}{(100)(9.8)} \]

Using the unrounded inner speed gives:

\[ \tan\theta_{\text{outer}} = \frac{222.89}{980} \]

\[ \tan\theta_{\text{outer}}=0.2274 \]

Therefore:

\[ \theta_{\text{outer}} = \tan^{-1}(0.2274) \]

\[ \theta_{\text{outer}}=12.8^\circ \]

The outer lane needs a banking angle of approximately \(12.8^\circ\).

Step 5: Explain why the outer lane is less steep.

This is the most important part of the question.

A tempting prediction is that the outer lane needs a steeper bank because it is further from the centre of the circular track.

But consider the required centripetal acceleration:

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

The car’s speed stays constant, while its radius increases from \(70.0\ \text{m}\) to \(100\ \text{m}\).

Therefore, its required centripetal acceleration decreases.

A smaller inward acceleration requires a smaller inward force relative to the vertical support force.

From the frictionless design-speed relationship:

\[ \frac{F_r}{mg}=\tan\theta \]

The larger radius therefore requires a smaller value of \(\tan\theta\) if the speed is to remain unchanged.

That is why the outer lane needs a smaller banking angle.

This is the key distinction: a steeper bank supports a higher design speed at a fixed radius, but a larger radius can support the same design speed with a shallower bank.

09Where this connects next

The next useful comparison is a flat bend.

On a flat road, the normal force acts vertically upwards, so it has no horizontal inward component. Tyre friction must provide the entire centripetal force.

Understanding that contrast makes it much easier to decide which real force supplies the radial acceleration in different circular-motion problems.

See How to Calculate the Maximum Speed on a Flat Bend for the next step.