How to Calculate the Maximum Speed on a Flat Bend

A car can travel around a flat bend at 45 km/h and lose grip at 50 km/h, even though the road itself hasn’t changed. The limiting factor is whether friction between the tyres and road can provide enough inward force to keep the car moving in a circle.

Suppose you double the radius of the bend but keep the tyres and road surface the same. Does the maximum speed double too? No. It increases by a factor of \(\sqrt{2}\), because the required centripetal acceleration depends on the square of the speed.

For a flat, level bend, the theoretical maximum speed before the tyres lose grip is:

\[ \boxed{v_{\max}=\sqrt{\mu_sgr}} \]

Here:

• \(v_{\max}\) is the maximum speed in \(\text{m s}^{-1}\).
• \(\mu_s\) is the coefficient of static friction between the tyres and road.
• \(g\) is gravitational acceleration, approximately \(9.8\text{ m s}^{-2}\).
• \(r\) is the radius of the car’s circular path, measured in metres.

This article shows where the formula comes from, how to estimate the friction coefficient experimentally, and how to use both pieces of information to calculate the maximum speed.

One important qualification: this is an idealised physics limit, not a recommended real-world driving speed. Actual road safety requires additional considerations.

01Identify the real force that turns the car

Before using the formula, you need to identify which physical force actually produces the circular motion.

Imagine a car travelling around a flat, circular bend at constant speed.

Its speed stays the same, but its velocity changes because its direction changes continuously. This means the car must accelerate towards the centre of the circular path.

That acceleration is called centripetal acceleration:

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

where \(a_c\) is measured in \(\text{m s}^{-2}\).

But what force produces this acceleration?

On a flat road, the answer is static friction between the tyres and the road.

There are three forces acting on the car in the simplified model:

The normal reaction and weight balance because the car has no vertical acceleration.

The only horizontal force in this model is static friction. It therefore provides the entire inward force required for circular motion.

This is an important distinction: centripetal force is not an additional physical force. It describes the net inward force acting on an object.

For a car on a flat bend, static friction is the real force supplying that centripetal force.

If you want to explore the underlying force diagrams and the difference between static and kinetic friction in more detail, see Cars on Flat Circular Bends: Friction and Centripetal Force.

02Derive the maximum speed

Now we can connect the friction force to the car’s speed.

Take the inward radial direction as positive and the vertically upwards direction as positive.

Because the car has no vertical acceleration, Newton’s second law gives:

\[ \sum F_y=N-mg=0 \]

Therefore:

\[ N=mg \]

In the radial direction, static friction provides the required centripetal force:

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

The important detail is that static friction has a maximum value.

It adjusts to the force required to maintain the circular path, provided that the required force does not exceed its limiting value:

\[ f_s\leq f_{s,\max} \]

The maximum static friction is:

\[ f_{s,\max}=\mu_sN \]

Substituting \(N=mg\):

\[ f_{s,\max}=\mu_smg \]

At the maximum speed, the tyres are just about to lose their grip. Static friction has reached its limiting value.

Therefore:

\[ \frac{mv_{\max}^2}{r}=\mu_smg \]

Cancel the mass \(m\):

\[ \frac{v_{\max}^2}{r}=\mu_sg \]

Rearrange to make speed the subject:

\[ v_{\max}^2=\mu_sgr \]

Hence:

\[ \boxed{v_{\max}=\sqrt{\mu_sgr}} \]

That’s the required relationship.

What does the formula tell us?

There are three useful predictions.

The mass cancellation makes sense physically.

A heavier car needs more inward force to maintain the same circular motion. However, its greater weight also increases the normal reaction and therefore the maximum available friction.

Both effects increase proportionally.

This result depends on the simplified friction model. Real tyre behaviour is more complicated.

The important distinction: actual friction versus maximum friction

A common mistake is to assume that:

\[ f_s=\mu_sN \]

at every speed.

That is incorrect.

The equality only applies when static friction reaches its maximum, immediately before slipping.

At lower speeds:

\[ f_s<\mu_sN \]

Think of \(\mu_sN\) as a ceiling on the available friction, rather than the amount of friction the tyres always produce.

For example, consider a \(1200\text{ kg}\) car travelling at \(10\text{ m s}^{-1}\) around a flat bend of radius \(50\text{ m}\). The coefficient of static friction is \(0.50\).

The actual friction required is:

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

\[ f_s=\frac{1200(10)^2}{50} =2400\text{ N} \]

However, the maximum available friction is:

\[ f_{s,\max}=\mu_smg \]

\[ f_{s,\max}=0.50(1200)(9.8) =5880\text{ N} \]

The car only requires \(2400\text{ N}\), even though the road could theoretically provide up to \(5880\text{ N}\).

The tyres are not at their friction limit.

03How can we determine the friction coefficient experimentally?

The formula is useful, but there’s a practical problem.

How do we know the coefficient of static friction between two surfaces?

It isn’t automatically known. It needs to be measured or supplied in the question.

A simple model investigation uses a rubber block, a horizontal surface and a force sensor.

Setting up the investigation

Place a rubber block on a horizontal sample of road surface.

Attach the block to a digital force sensor using a light string. The string should be horizontal, so the pulling force does not change the normal reaction.

The force sensor records the horizontal force applied to the block.

Gradually increase the pulling force.

Initially, the block remains stationary. Static friction opposes the pulling force.

Eventually, the applied force reaches the maximum static friction and the block begins to slide.

The highest force reading immediately before sliding provides an estimate of the limiting static friction.

Why not simply perform this experiment once?

A single measurement gives one estimate of the coefficient, but it doesn’t test whether the friction force is proportional to the normal force.

A better investigation repeats the procedure using several different normal forces.

For each trial:

1. Change the mass on top of the block.
2. Calculate the resulting normal reaction.
3. Increase the pulling force slowly.
4. Record the maximum force immediately before sliding.
5. Repeat the measurement and calculate a mean.

The prediction is that the maximum static friction is proportional to the normal reaction:

\[ F_{s,\max}=\mu_sN \]

Plotting maximum static friction against normal force should produce an approximately straight line.

The gradient represents the coefficient of static friction.

Interpreting the results

Suppose the investigation produces the following illustrative data.

The readings increase approximately proportionally.

The gradient of the best-fit line is approximately:

\[ \mu_s\approx0.405 \]

The coefficient of static friction is dimensionless because it is the ratio of two forces measured in newtons.

This investigation supports the limiting-friction model for the tested surfaces.

However, there is an important limitation.

A rubber block is not a complete model of a car tyre. Actual tyre-road friction depends on tyre construction, tread, temperature, surface condition, water, and other factors.

The experiment estimates the coefficient for the particular sample and conditions tested. It does not directly establish the safe speed of an actual road vehicle.

The data above are illustrative, not measurements from a real road experiment.

04Worked examples

05When should you use a different equation?

The equation

\[ v_{\max}=\sqrt{\mu_sgr} \]

is specifically for a flat, level bend with constant-speed circular motion.

It assumes:

• The road is horizontal, not banked.
• Static friction supplies the entire horizontal centripetal force.
• The car is not braking or accelerating tangentially.
• The coefficient of static friction is constant.
• No other horizontal forces contribute significantly.

Changing these conditions can change the force analysis.

Flat bend versus banked bend

A banked road slopes upwards towards the outside of the bend.

This changes the direction of the normal reaction. Instead of acting vertically upwards, it is tilted towards the centre of the circular path.

Consequently, the normal reaction can now provide an inward component of force.

For a flat road, static friction supplies the entire inward force.

For an ideal frictionless banked road, the horizontal component of the normal reaction supplies the centripetal force.

These are different force models, so the flat-bend equation cannot simply be transferred to a banked road.

For the separate banked-road calculation, see How to Calculate the Design Speed of a Frictionless Banked Road.

Maximum theoretical speed versus recommended speed

Another important distinction is the difference between the maximum speed predicted by a friction model and a recommended road speed.

The equation calculates the theoretical point at which the available static friction reaches its limit.

It does not account for every factor affecting real driving, such as:

• Changing road conditions.
• Tyre wear and temperature.
• Visibility and reaction time.
• Variations in road curvature.
• Vehicle handling and suspension.
• Appropriate safety margins.

A posted speed limit is not necessarily the speed at which the tyres would begin to slide.

Do not use this equation to choose a real-world driving speed.

06NSW HSC syllabus note

For students completing the 2017 Physics Stage 6 syllabus, circular motion is part of the Year 12 course. Its Circular Motion content explicitly includes analysing cars moving around horizontal circular bends.

The newer Physics 11-12 Syllabus (2025) begins implementation for Year 11 in 2027 and Year 12 in Term 4 of 2027. Its first HSC examination is in 2028.

This resource develops the horizontal-bend application in the 2017 syllabus. Students transitioning to the newer course should check the precise content and depth required in their school’s program.

See NESA’s Physics Stage 6 syllabus information.

07Questions and solutions

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

Question 1

A \(1200\text{ kg}\) car travels around a flat, level bend with a radius of \(48\text{ m}\).

The coefficient of static friction between the tyres and road is \(0.50\).

Calculate:

a. The maximum speed before the car loses grip, in \(\text{m s}^{-1}\).

b. The maximum speed in \(\text{km h}^{-1}\).

c. Identify the real force providing the centripetal force.

Solution 1

The maximum speed is approximately \(15.3\text{ m s}^{-1}\), or \(55.2\text{ km h}^{-1}\). Static friction between the tyres and road provides the centripetal force.

Because the road is flat and the car travels at constant speed, we can use:

\[ v_{\max}=\sqrt{\mu_sgr} \]

Substitute the values:

\[ \begin{aligned} v_{\max} &=\sqrt{(0.50)(9.8)(48)}\\ &=\sqrt{235.2}\\ &=15.34\text{ m s}^{-1} \end{aligned} \]

Convert to kilometres per hour:

\[ v_{\max}=15.34\times3.6 \]

\[ v_{\max}=55.2\text{ km h}^{-1} \]

For part c, static friction acts horizontally towards the centre of the bend.

The car’s mass is not needed because it cancels when the centripetal force is equated to the maximum static friction.

The important point is that we have not added a separate centripetal force to the free-body diagram. Friction is the real force providing the required inward acceleration.

Question 2

A road-testing team uses a rubber sample to estimate the coefficient of static friction for a particular road surface.

The following results are obtained.

For this question, assume the sample provides a suitable model of the tyre-road contact.

A car travels around a flat, level bend with a radius of \(62\text{ m}\).

a. Estimate the coefficient of static friction using the gradient of a best-fit line through the origin.

b. Calculate the theoretical maximum speed.

c. The car is actually travelling at \(50\text{ km h}^{-1}\). Determine whether the model predicts that it will slip.

d. Calculate approximately what percentage of the maximum available static friction is being used at \(50\text{ km h}^{-1}\).

Solution 2

The estimated coefficient of static friction is approximately \(0.42\), giving a maximum speed of \(16.0\text{ m s}^{-1}\), or \(57.5\text{ km h}^{-1}\).

At \(50\text{ km h}^{-1}\), the car uses approximately \(76\%\) of its available static friction, so the model predicts that it will not slip.

Part a: Estimate the coefficient of friction.

The coefficient is the gradient of the graph:

\[ \mu_s=\frac{\Delta F_{s,\max}}{\Delta N} \]

Using the first and last readings:

\[ \mu_s\approx\frac{6.7-1.7}{16-4} \]

\[ \mu_s\approx\frac{5.0}{12} \]

\[ \mu_s\approx0.417 \]

Therefore:

\[ \mu_s\approx0.42 \]

Part b: Calculate the maximum speed.

Use:

\[ v_{\max}=\sqrt{\mu_sgr} \]

Substitute:

\[ v_{\max}=\sqrt{(0.42)(9.8)(62)} \]

\[ v_{\max}=\sqrt{255.192} \]

\[ v_{\max}=15.97\text{ m s}^{-1} \]

Convert to kilometres per hour:

\[ v_{\max}=15.97\times3.6 \]

\[ v_{\max}=57.5\text{ km h}^{-1} \]

Part c: Determine whether the car slips at 50 km/h.

First, convert the actual speed:

\[ v=\frac{50}{3.6}=13.89\text{ m s}^{-1} \]

Calculate the required centripetal acceleration:

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

\[ a_c=\frac{13.89^2}{62} \]

\[ a_c=3.11\text{ m s}^{-2} \]

The maximum available frictional acceleration is:

\[ a_{\max}=\mu_sg \]

\[ a_{\max}=(0.42)(9.8) \]

\[ a_{\max}=4.12\text{ m s}^{-2} \]

Since:

\[ 3.11<4.12 \]

the required inward acceleration is below the maximum available acceleration.

The car can therefore maintain the circular path without slipping according to the model.

Part d: Calculate the percentage of available friction being used.

\[ \text{Percentage used} =\frac{a_c}{a_{\max}}\times100 \]

\[ =\frac{3.11}{4.12}\times100 \]

\[ =75.6\% \]

Approximately \(76\%\) of the available static friction is being used.

The tempting mistake is to assume the friction must already equal \(\mu_sN\) because the car is moving around a bend.

It doesn’t. Static friction only needs to provide the force required at the current speed.

Question 3

A student investigates the coefficient of static friction between a rubber block and a horizontal surface.

The block has a mass of \(2.5\text{ kg}\).

A force sensor pulls the block at an angle of \(25^\circ\) above the horizontal. Immediately before the block begins sliding, the sensor records a force of \(10.0\text{ N}\).

Assume the block has no vertical acceleration.

a. Calculate the normal reaction force.

b. Determine the maximum static friction force.

c. Calculate the coefficient of static friction.

d. Explain why calculating \(\mu_s=F/(mg)\) would give an incorrect result.

Solution 3

The coefficient of static friction is approximately \(0.45\). The normal reaction is approximately \(20.3\text{ N}\), and the maximum static friction is approximately \(9.06\text{ N}\).

The key is that the pulling force has both horizontal and vertical components.

Part a: Calculate the normal reaction.

There is no vertical acceleration.

Taking upwards as positive:

\[ N+F\sin25^\circ-mg=0 \]

Rearrange:

\[ N=mg-F\sin25^\circ \]

Substitute:

\[ N=(2.5)(9.8)-(10.0)\sin25^\circ \]

\[ N=24.5-4.23 \]

\[ N=20.27\text{ N} \]

Part b: Calculate the maximum static friction.

Immediately before sliding, the horizontal forces balance.

Therefore, the maximum static friction equals the horizontal component of the pulling force:

\[ f_{s,\max}=F\cos25^\circ \]

\[ f_{s,\max}=10.0\cos25^\circ \]

\[ f_{s,\max}=9.06\text{ N} \]

Part c: Calculate the coefficient.

Use:

\[ f_{s,\max}=\mu_sN \]

Rearrange:

\[ \mu_s=\frac{f_{s,\max}}{N} \]

Substitute:

\[ \mu_s=\frac{9.06}{20.27} \]

\[ \mu_s=0.447 \]

Therefore:

\[ \boxed{\mu_s\approx0.45} \]

Part d: Explain the incorrect calculation.

Using \(F/(mg)\) incorrectly treats the entire pulling force as horizontal and assumes the normal reaction equals the block’s weight.

Neither assumption is correct.

The upward component of the pulling force reduces the normal reaction, while only the horizontal component opposes friction.

This is why the coefficient must be calculated using the resolved forces.

Question 4

A car travels around a banked circular bend at a constant speed of \(18.0\text{ m s}^{-1}\).

The road is banked at \(12^\circ\) to the horizontal.

Assume friction is negligible at the design speed.

a. Calculate the radius of the bend.

b. Identify the real force providing the centripetal force.

c. Explain why the flat-bend equation \(v_{\max}=\sqrt{\mu_sgr}\) is inappropriate.

Solution 4

The bend has a radius of approximately \(156\text{ m}\). The horizontal component of the normal reaction supplies the centripetal force.

This is different from a flat bend because the normal reaction is tilted.

Part a: Calculate the radius.

Resolve the normal reaction into vertical and horizontal components.

Vertically, the forces balance:

\[ N\cos12^\circ=mg \]

Horizontally, the inward component supplies the centripetal force:

\[ N\sin12^\circ=\frac{mv^2}{r} \]

Divide the horizontal equation by the vertical equation:

\[ \frac{N\sin12^\circ}{N\cos12^\circ} = \frac{mv^2/r}{mg} \]

Simplify:

\[ \tan12^\circ=\frac{v^2}{rg} \]

Rearrange:

\[ r=\frac{v^2}{g\tan12^\circ} \]

Substitute:

\[ r=\frac{18.0^2}{9.8\tan12^\circ} \]

\[ r=\frac{324}{2.082} \]

\[ r=155.6\text{ m} \]

Therefore:

\[ r\approx156\text{ m} \]

Part b: Identify the real centripetal force.

The horizontal component of the normal reaction acts towards the centre of the circular path.

It provides the required centripetal force.

Part c: Explain why the flat-bend equation is inappropriate.

The flat-bend equation assumes that the normal reaction is vertical and balances the weight:

\[ N=mg \]

It also assumes static friction supplies the entire inward force.

On a banked road, the normal reaction has a horizontal component.

The force model is therefore different, so the flat-bend equation cannot be used directly.

Question 5

Extension: A car with speed-dependent aerodynamic downforce

A \(900\text{ kg}\) experimental car travels around a flat, circular test bend with a radius of \(45\text{ m}\).

The coefficient of static friction between its tyres and the road is \(0.80\).

An aerodynamic device produces an additional downward force of \(3.0\text{ kN}\).

For the first part of this question, assume:

• The downforce is constant at the instant being considered.
• The road is flat and level.
• The coefficient of static friction remains constant.
• There is no braking or tangential acceleration.
• There are no other horizontal forces.

a. Calculate the normal reaction.

b. Calculate the maximum available static friction.

c. Determine the maximum speed before slipping.

d. Compare this result with the maximum speed if the car had no aerodynamic downforce.

For a separate idealised model, suppose the downforce instead increases with speed according to:

\[ D=kv^2 \]

where:

\[ k=28\text{ N s}^2\text{ m}^{-2} \]

e. Determine whether this second model predicts a finite maximum speed due to tyre-road friction. Justify your conclusion mathematically.

Solution 5

With a constant downforce of \(3.0\text{ kN}\), the model predicts a maximum speed of approximately \(21.7\text{ m s}^{-1}\), or \(78.3\text{ km h}^{-1}\).

Without downforce, the maximum speed is approximately \(18.8\text{ m s}^{-1}\), or \(67.6\text{ km h}^{-1}\).

For the second model, no finite friction-limited speed is predicted.

The interesting feature is that aerodynamic downforce increases the normal reaction, which increases the maximum available static friction.

Part a: Calculate the normal reaction.

There are now three vertical forces:

• Normal reaction upwards.
• Weight downwards.
• Aerodynamic downforce downwards.

There is no vertical acceleration, so:

\[ N-mg-D=0 \]

Rearrange:

\[ N=mg+D \]

Substitute:

\[ N=(900)(9.8)+3000 \]

\[ N=8820+3000 \]

\[ N=11820\text{ N} \]

Part b: Calculate the maximum static friction.

Use:

\[ f_{s,\max}=\mu_sN \]

Substitute:

\[ f_{s,\max}=(0.80)(11820) \]

\[ f_{s,\max}=9456\text{ N} \]

Part c: Calculate the maximum speed.

At the limiting speed, static friction supplies the entire centripetal force:

\[ f_{s,\max}=\frac{mv_{\max}^2}{r} \]

Rearrange:

\[ v_{\max}=\sqrt{\frac{f_{s,\max}r}{m}} \]

Substitute:

\[ v_{\max}=\sqrt{\frac{(9456)(45)}{900}} \]

\[ v_{\max}=\sqrt{472.8} \]

\[ v_{\max}=21.74\text{ m s}^{-1} \]

Convert:

\[ v_{\max}=21.74\times3.6 \]

\[ v_{\max}=78.3\text{ km h}^{-1} \]

Part d: Compare with no downforce.

Without downforce:

\[ N=mg \]

The ordinary flat-bend equation applies:

\[ v_{\max}=\sqrt{\mu_sgr} \]

Substitute:

\[ v_{\max}=\sqrt{(0.80)(9.8)(45)} \]

\[ v_{\max}=\sqrt{352.8} \]

\[ v_{\max}=18.78\text{ m s}^{-1} \]

Convert:

\[ v_{\max}=18.78\times3.6 \]

\[ v_{\max}=67.6\text{ km h}^{-1} \]

The downforce increases the theoretical maximum speed from approximately \(67.6\text{ km h}^{-1}\) to \(78.3\text{ km h}^{-1}\).

The tempting mistake is to use \(N=mg\) automatically and conclude that downforce makes no difference because mass cancels from the ordinary equation.

But the normal reaction is now greater than the car’s weight. The maximum friction increases, even though the car’s mass has not changed.

Part e: Analyse the speed-dependent downforce model.

Now replace the constant downforce with:

\[ D=kv^2 \]

The normal reaction becomes:

\[ N=mg+kv^2 \]

The maximum available static friction is:

\[ f_{s,\max}=\mu_s(mg+kv^2) \]

The required centripetal force is still:

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

For the car to maintain circular motion without slipping:

\[ \frac{mv^2}{r}\leq\mu_s(mg+kv^2) \]

Expand:

\[ \frac{mv^2}{r}\leq\mu_smg+\mu_skv^2 \]

Collect the terms containing \(v^2\):

\[ \left(\frac{m}{r}-\mu_sk\right)v^2 \leq\mu_smg \]

Now compare the two coefficients.

\[ \frac{m}{r}=\frac{900}{45}=20\text{ kg m}^{-1} \]

And:

\[ \mu_sk=(0.80)(28) =22.4\text{ kg m}^{-1} \]

Therefore:

\[ (20-22.4)v^2\leq(0.80)(900)(9.8) \]

\[ -2.4v^2\leq7056 \]

For every real speed, \(v^2\) is non-negative. The left-hand side is therefore zero or negative, while the right-hand side is positive.

The inequality is satisfied at every finite speed.

The simplified model consequently predicts no finite maximum speed imposed by tyre-road friction.

This is the important conceptual result: the required centripetal force increases with \(v^2\), but the available friction also increases with \(v^2\) because the downforce increases the normal reaction. In this particular model, the available friction increases faster.

It does not mean an actual car could travel at unlimited speed. The model ignores aerodynamic drag, tyre behaviour at extreme speeds, structural limits, and other real-world constraints.

The extension illustrates why you must check the vertical forces before assuming \(N=mg\).

For ordinary HSC flat-bend problems, return to the central decision: identify the real inward force, determine the maximum available friction, and equate the two only at the limiting speed. If the road is banked or another force changes the normal reaction, draw the full force diagram before selecting an equation.