Why Centripetal Force Is Not an Extra Force in HSC Physics

At the bottom of a vertical loop, a cart experiences an upward normal force from the track and a downward gravitational force. A student draws both arrows, then adds a third upward arrow labelled centripetal force.

The diagram now looks more complete. Unfortunately, it counts the inward effect twice.

You should not add a separate centripetal-force arrow to a free-body diagram. Centripetal force is not an additional physical interaction. It is the name given to the net inward radial force produced by the actual forces acting on the object.

Draw the real forces first. Then calculate their resultant towards the centre of the circular path.

Before going further, make a prediction. Suppose the track pushes upwards with a force of \(12\text{ N}\), while gravity pulls downwards with a force of \(4\text{ N}\). What is the net inward force?

It is \(8\text{ N}\), not \(12\text{ N}\) or \(16\text{ N}\). The centripetal force is that \(8\text{ N}\) resultant. It is not a third force.

01A free-body diagram shows actual forces, not the required acceleration

A free-body diagram isolates an object and shows the forces exerted on it by other objects or fields.

For a cart moving around a vertical circular track, these might include:

• Weight, \(F_g=mg\), acting vertically downwards.
• Normal force, \(N\), exerted by the track.
• Friction, if the track is not perfectly smooth.

Each force represents a real interaction.

Centripetal force represents something different: the combined effect of the forces in the inward radial direction.

A useful analogy is a shopping receipt.

You list the individual purchases, then calculate the subtotal. You wouldn’t add the subtotal as another purchase. That would count the same money twice.

The actual forces are the individual purchases. The centripetal force is the subtotal of their inward radial components.

There is one important difference: forces are vectors. Their directions matter, so you must add their components with the correct signs.

Can you ever draw an arrow labelled centripetal force?

Yes, provided you understand what the arrow represents.

You can draw an arrow showing the net inward radial force on an object. You might label this arrow \(F_c\).

However, it must represent the resultant of the actual forces, not an additional force to include alongside them.

For example, if a ball is attached to a horizontal string and moves in a circle, the string tension might be the only horizontal force acting on it.

In that situation:

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

Tension is the actual physical force. Centripetal force describes the role that tension plays in producing circular motion.

They are not two separate forces.

02Radial force and tangential speed are different things

To understand circular-motion force diagrams properly, you need to separate two directions.

Consider an object travelling around a circular path.

Its velocity points along the tangent to the circle. This is the direction in which the object is moving at that instant.

The radial direction points towards the centre of the circle.

These directions are always perpendicular.

Radial acceleration changes direction

Even if an object travels at constant speed, its velocity changes because its direction changes.

This requires an acceleration towards the centre of the circle.

Its magnitude is:

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

where:

• \(a_r\) is radial (centripetal) acceleration, in \(\text{m s}^{-2}\).
• \(v\) is tangential speed, in \(\text{m s}^{-1}\).
• \(r\) is the radius of the circular path, in metres.

Using Newton’s second law, the net inward radial force must therefore be:

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

Here, \(m\) is the object’s mass in kilograms.

The left-hand side is the sum of the radial components of the actual forces. It is not a sum that includes an additional centripetal force.

For the equations in this article, take the inward radial direction as positive.

Tangential acceleration changes speed

Now consider what happens if the object speeds up or slows down while travelling around the circle.

Its speed changes, so it must have a tangential acceleration.

The tangential acceleration is:

\[
a_t=\frac{dv}{dt}
\]

The corresponding force equation is:

\[
\boxed{\sum F_t=ma_t}
\]

where \(a_t\) is tangential acceleration, measured in \(\text{m s}^{-2}\).

If the object speeds up, tangential acceleration points in the direction of motion.

If it slows down, tangential acceleration points opposite to its motion.

This gives us two separate equations:

For uniform circular motion, speed remains constant, so tangential acceleration is zero.

For non-uniform circular motion, both radial and tangential acceleration can exist simultaneously.

This is an important distinction. The expression \(mv^2/r\) gives the radial component of the net force. It does not necessarily give the magnitude of the total net force.

When both components are present:

\[
F_{\text{net}}=
\sqrt{
\left(\frac{mv^2}{r}\right)^2+
(ma_t)^2
}
\]

The two components are perpendicular, which is why Pythagoras applies.

03A vertical loop shows why the force diagram changes with position

Let’s use a cart travelling around the inside of a smooth vertical circular track.

The cart experiences two real forces:

• Normal force, \(N\), from the track.
• Weight, \(mg\), from gravity.

The forces are real throughout the motion. However, their directions relative to the radial axis change as the cart moves.

Let’s examine each position.

At the bottom of the loop

The centre of the circle is above the cart.

Therefore, the inward radial direction is upwards.

• Normal force acts upwards.
• Weight acts downwards.

Taking upwards as positive:

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

Rearranging:

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

Notice that the normal force must be greater than the weight.

The difference between these forces supplies the required inward resultant.

You should not add another upward centripetal-force arrow.

At the top of the loop

The centre of the circle is now below the cart.

The inward radial direction is downwards.

Both normal force and gravity point towards the centre.

Therefore:

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

Rearranging:

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

Gravity now contributes to the required inward resultant rather than opposing it.

This is why you cannot automatically use the same force equation at the top and bottom of a vertical loop.

At the side of the loop

At the right-hand side, the centre is to the left.

The normal force points horizontally towards the centre.

Gravity points vertically downwards.

If the cart is moving upwards, gravity acts opposite to its instantaneous velocity. It therefore slows the cart down.

The radial equation is:

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

The tangential equation is:

\[
mg=ma_t
\]

with downward taken as positive.

So:

\[
a_t=g
\]

The normal force changes the direction of the velocity, while gravity changes its speed at this instant.

This is a particularly useful example because it shows that a real force can be radial at one point in a circular path and tangential at another.

04An experiment: measuring the actual force in a circular path

How could we test whether a force sensor measures the centripetal force directly, or one of the actual forces contributing to it?

Consider a proposed investigation using a small cart moving around the inside of a vertical circular track.

This is an illustrative experimental design, not a report of a published experimental data set.

Setting up the apparatus

Use the following equipment:

• A smooth vertical circular track with a radius of \(1.00\text{ m}\).
• A small cart with a mass of \(0.20\text{ kg}\).
• A calibrated force sensor built into a short section of track at the bottom.
• A photogate positioned immediately beside the force-sensing section.
• A small flag of known length attached to the cart.
• A ruler to measure the track’s radius.

The force sensor measures the contact force exerted by the cart on the track. By Newton’s third law, this has the same magnitude as the normal force exerted by the track on the cart.

The photogate measures the cart’s instantaneous speed as its flag passes through the beam.

What should the experiment show?

At the bottom of the track, the inward direction is upwards.

The normal force acts upwards, while gravity acts downwards.

Therefore, our prediction is:

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

Rearranging:

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

This predicts that the sensor’s reading should be greater than the required net inward force by exactly the object’s weight.

The problem with measuring only the contact force

Imagine that the force sensor records \(14.8\text{ N}\).

A student might call this the centripetal force.

But the sensor measures the normal force, not the net radial force.

Without measuring the cart’s mass, speed, and radius, we cannot establish the radial resultant from this reading alone.

We need to revise the investigation by adding independent measurements of these quantities.

Revising the investigation

Suppose the following illustrative measurements are obtained:

Assume the cart remains in contact with the track, and friction and air resistance are negligible.

We can now calculate the expected net inward force.

Step 1: Calculate the required radial resultant.

\[
\begin{aligned}
F_{r,\text{net}}&=\frac{mv^2}{r}\\
&=\frac{0.20\times 8.0^2}{1.00}\\
&=12.8\text{ N}
\end{aligned}
\]

Step 2: Calculate the cart’s weight.

\[
\begin{aligned}
mg&=0.20\times9.8\\
&=1.96\text{ N}
\end{aligned}
\]

Step 3: Calculate the predicted normal force.

\[
\begin{aligned}
N&=mg+F_{r,\text{net}}\\
&=1.96+12.8\\
&=14.76\text{ N}
\end{aligned}
\]

Therefore:

\[
N\approx14.8\text{ N}
\]

This agrees with the illustrative sensor reading.

We can also work backwards:

\[
F_{r,\text{net}}=14.8-1.96
\]

\[
F_{r,\text{net}}=12.84\text{ N}
\]

The measured radial resultant is approximately \(12.8\text{ N}\).

What did the revised experiment establish?

The force sensor measured the normal force.

The net inward radial force was the difference between the normal force and weight.

The experiment therefore supports the prediction that the centripetal force is the resultant of the actual forces, not an additional force.

There are also practical limitations to consider:

• A stopwatch measuring a complete revolution gives an average speed, not necessarily the instantaneous speed at the bottom.
• Friction and air resistance can change the cart’s speed.
• The track’s radius must be measured accurately.
• The force sensor must be calibrated without significantly changing the track’s behaviour.

Using a photogate beside the force sensor addresses one important limitation: the speed and contact force are measured at approximately the same position.

05Worked examples

06What changes when you use a rotating reference frame?

So far, we have described circular motion from the perspective of an observer standing on the ground.

This is an approximately inertial reference frame, meaning Newton’s laws can be applied without introducing fictitious forces.

But what happens if you sit on a rotating carousel?

Imagine a small object resting on the carousel at a fixed distance from its centre.

A person standing on the ground sees the object travelling in a circle.

A person rotating with the carousel sees the object remaining stationary relative to them.

Both descriptions can be correct.

From the ground observer’s perspective

The object moves in a circle.

Its velocity changes direction, so it has inward acceleration.

The real static friction force provides the required inward resultant:

\[
F_{\text{friction}}=\frac{mv^2}{r}
\]

The free-body diagram contains the actual forces:

• Weight downwards.
• Normal force upwards.
• Static friction towards the centre.

There is no centrifugal force in this inertial-frame description.

There is also no additional centripetal force.

From the rotating observer’s perspective

The object appears stationary relative to the carousel.

However, the real friction force still acts towards the centre.

To apply Newton’s laws in this rotating, non-inertial reference frame, we introduce an apparent outward force called the centrifugal force.

This is a pseudo-force. It is not caused by a physical interaction with another object.

For an object stationary relative to a uniformly rotating frame, the outward centrifugal pseudo-force balances the real inward force.

Its magnitude is:

\[
F_{\text{centrifugal}}=\frac{mv^2}{r}
\]

The two descriptions are consistent:

The important rule is to avoid mixing these descriptions.

If you use the ground frame, draw the actual forces and calculate their inward resultant.

If you use a rotating frame, introduce the appropriate pseudo-forces to account for the non-inertial reference frame.

Do not add a centripetal force and a centrifugal force to an ordinary ground-frame free-body diagram.

For a more detailed comparison of these two descriptions, see Centripetal vs Centrifugal Force: The Key Differences.

07A reliable method for circular-motion force diagrams

When approaching an HSC circular-motion problem, use the following sequence.

Step 1: Identify the object.

Isolate the object you are analysing. Do not include forces acting on other objects.

Step 2: Choose your reference frame.

Unless the question specifies otherwise, use the ground or laboratory frame.

Step 3: Draw the actual forces.

Include gravity, normal forces, tension, friction, and other real interactions where relevant.

Do not add centripetal force as an independent interaction.

Step 4: Identify the radial direction.

Draw an imaginary line from the object towards the centre of the circular path.

Take inward as positive unless you have a reason to choose another sign convention.

Step 5: Resolve the forces.

Calculate the sum of their radial components:

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

Then consider the tangential components separately:

\[
\sum F_t=ma_t
\]

Step 6: Check for double counting.

Ask yourself whether you have included a resultant force alongside the actual forces that already produce it.

If so, remove the extra centripetal-force term.

Common mistakes and how to avoid them

08Questions and solutions

Question 1

A \(0.30\text{ kg}\) stopper is attached to a light horizontal string and moves in a horizontal circle of radius \(0.80\text{ m}\) at a constant speed of \(4.0\text{ m s}^{-1}\).

Assume the table is smooth and air resistance is negligible.

(a) Calculate the tension in the string.

(b) Explain whether centripetal force should be drawn as a separate force on the stopper’s free-body diagram.

Solution 1

The string tension is \(6.0\text{ N}\) towards the centre. Centripetal force should not be included as an additional force.

Step 1: Identify the actual horizontal force.

The string tension acts towards the centre of the circle.

The normal force and weight balance vertically, so tension supplies the entire horizontal inward resultant.

Step 2: Apply the radial force equation.

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

Substitute:

\[
\begin{aligned}
T&=\frac{0.30\times4.0^2}{0.80}\\
&=\frac{4.8}{0.80}\\
&=6.0\text{ N}
\end{aligned}
\]

Step 3: Explain the force diagram.

The free-body diagram contains:

• Tension towards the centre.
• Weight downwards.
• Normal force upwards.

Tension already provides the required centripetal force.

Drawing another \(6.0\text{ N}\) inward arrow and adding it to tension would double-count the same effect.

Question 2

A \(0.25\text{ kg}\) cart travels around the inside of a smooth vertical circular track with a radius of \(1.0\text{ m}\).

At the bottom of the track, its speed is \(8.0\text{ m s}^{-1}\).

Assume negligible energy losses and take \(g=9.8\text{ m s}^{-2}\).

Calculate:

(a) The normal force acting on the cart at the bottom.

(b) The speed of the cart at the top.

(c) The normal force acting on the cart at the top.

(d) Whether the cart remains in contact with the track at the top.

Solution 2

The normal force is \(18.45\text{ N}\) at the bottom and \(3.75\text{ N}\) at the top. The cart remains in contact with the track because the normal force at the top is positive.

Step 1: Calculate the normal force at the bottom.

At the bottom, inward is upwards.

Normal force acts upwards, while weight acts downwards.

Therefore:

\[
N_b-mg=\frac{mv_b^2}{r}
\]

Rearranging:

\[
N_b=m\left(\frac{v_b^2}{r}+g\right)
\]

Substitute:

\[
\begin{aligned}
N_b&=0.25\left(\frac{8.0^2}{1.0}+9.8\right)\\
&=0.25(64+9.8)\\
&=18.45\text{ N}
\end{aligned}
\]

Step 2: Calculate the speed at the top.

The cart gains gravitational potential energy as it rises through a vertical height of \(2r\).

Using conservation of mechanical energy:

\[
\frac12mv_b^2
=
\frac12mv_t^2+mg(2r)
\]

Rearrange:

\[
v_t^2=v_b^2-4gr
\]

Substitute:

\[
\begin{aligned}
v_t^2&=8.0^2-4(9.8)(1.0)\\
&=64-39.2\\
&=24.8
\end{aligned}
\]

Therefore:

\[
v_t=\sqrt{24.8}
\]

\[
\boxed{v_t=4.98\text{ m s}^{-1}}
\]

Step 3: Calculate the normal force at the top.

At the top, inward is downwards.

Both gravity and the normal force point towards the centre.

Therefore:

\[
N_t+mg=\frac{mv_t^2}{r}
\]

Rearranging:

\[
N_t=m\left(\frac{v_t^2}{r}-g\right)
\]

Substitute:

\[
\begin{aligned}
N_t&=0.25\left(\frac{24.8}{1.0}-9.8\right)\\
&=0.25(15.0)\\
&=3.75\text{ N}
\end{aligned}
\]

Step 4: Determine whether contact is maintained.

The normal force is positive, meaning the track continues to push on the cart.

Therefore, the cart remains in contact with the track at the top.

The important lesson is that the normal force changes because gravity contributes differently to the inward resultant at different positions.

Question 3

A \(0.20\text{ kg}\) cart moves around the inside of a smooth vertical circular track with a radius of \(0.80\text{ m}\).

At the right-hand side of the track, the cart is moving upwards at \(4.0\text{ m s}^{-1}\).

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

Determine:

(a) The normal force acting on the cart.

(b) Its tangential acceleration.

(c) The magnitude of its total acceleration.

Explain why \(mv^2/r\) does not equal the magnitude of the total net force in this situation.

Solution 3

The normal force is \(4.0\text{ N}\) towards the centre, the tangential acceleration is \(9.8\text{ m s}^{-2}\) downwards, and the total acceleration is approximately \(22.3\text{ m s}^{-2}\).

Step 1: Calculate the radial acceleration.

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

Substitute:

\[
\begin{aligned}
a_r&=\frac{4.0^2}{0.80}\\
&=\frac{16}{0.80}\\
&=20\text{ m s}^{-2}
\end{aligned}
\]

The radial acceleration is towards the centre.

Step 2: Calculate the normal force.

At the right-hand side, the normal force points horizontally towards the centre.

Gravity acts vertically downwards, so it has no radial component at this position.

Therefore:

\[
N=ma_r
\]

\[
N=0.20\times20
\]

\[
\boxed{N=4.0\text{ N}}
\]

Step 3: Calculate the tangential acceleration.

The cart is moving upwards.

Gravity acts downwards, opposite to the cart’s direction of motion.

Because the track is smooth, gravity provides the entire tangential resultant.

\[
ma_t=mg
\]

Therefore:

\[
\boxed{a_t=9.8\text{ m s}^{-2}\text{ downwards}}
\]

The cart is slowing down as it moves upwards.

Step 4: Calculate the total acceleration.

The radial and tangential acceleration components are perpendicular.

Therefore:

\[
a_{\text{net}}=\sqrt{a_r^2+a_t^2}
\]

Substitute:

\[
\begin{aligned}
a_{\text{net}}
&=\sqrt{20^2+9.8^2}\\
&=\sqrt{400+96.04}\\
&=\sqrt{496.04}\\
&=22.27\text{ m s}^{-2}
\end{aligned}
\]

Thus:

\[
\boxed{a_{\text{net}}\approx22.3\text{ m s}^{-2}}
\]

The radial acceleration is only one component of the total acceleration.

Similarly, the radial net force is:

\[
F_r=0.20\times20=4.0\text{ N}
\]

The tangential force is:

\[
F_t=0.20\times9.8=1.96\text{ N}
\]

The total net force is therefore:

\[
F_{\text{net}}=\sqrt{4.0^2+1.96^2}
\]

\[
\boxed{F_{\text{net}}=4.45\text{ N}}
\]

This question demonstrates why you must analyse radial and tangential components separately when an object’s speed changes.

Question 4

A \(0.50\text{ kg}\) block rests on a horizontal rotating carousel at a distance of \(2.0\text{ m}\) from its centre.

The carousel rotates at a constant period of \(4.0\text{ s}\), and the block does not slip.

(a) Calculate the magnitude of the static friction force acting on the block from the perspective of an observer standing on the ground.

(b) Describe the forces an observer rotating with the carousel would use to explain why the block remains stationary relative to the platform.

(c) Explain why the centrifugal force should not be added to the ground-frame free-body diagram.

Solution 4

The static friction force is approximately \(2.47\text{ N}\) towards the centre in the ground frame. In the rotating frame, an outward centrifugal pseudo-force of the same magnitude balances this inward friction force.

Step 1: Calculate the tangential speed.

The relationship between speed, radius and period is:

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

Substitute:

\[
\begin{aligned}
v&=\frac{2\pi(2.0)}{4.0}\\
&=\pi\\
&=3.14\text{ m s}^{-1}
\end{aligned}
\]

Step 2: Calculate the inward resultant force.

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

Substitute:

\[
\begin{aligned}
F_r&=\frac{0.50\times\pi^2}{2.0}\\
&=2.47\text{ N}
\end{aligned}
\]

The block requires a net inward force of \(2.47\text{ N}\).

In the ground frame, static friction supplies this force.

Step 3: Explain the rotating-frame description.

To an observer rotating with the carousel, the block appears stationary.

To apply Newton’s laws in this non-inertial frame, the observer introduces an outward centrifugal pseudo-force.

Its magnitude is:

\[
F_{\text{centrifugal}}=2.47\text{ N}
\]

This balances the real inward static friction force.

Step 4: Explain why the two forces cannot be mixed.

The ground observer does not require a centrifugal force because their reference frame is approximately inertial.

The rotating observer introduces the centrifugal pseudo-force because their reference frame is accelerating.

They are different descriptions of the same physical situation.

Question 5

A \(1000\text{ kg}\) car travels around a level circular test track with a radius of \(40\text{ m}\).

The coefficient of static friction between the tyres and road is \(0.50\). Take \(g=9.8\text{ m s}^{-2}\).

Assume:

• The road is level.
• There is no aerodynamic downforce.
• Air resistance is negligible.
• The combined horizontal tyre-road contact force is limited by the maximum static friction.

At a particular instant, the car travels at \(12\text{ m s}^{-1}\), and the driver requests a tangential acceleration of \(4.0\text{ m s}^{-2}\).

Two students disagree.

• Student A argues that the car should be able to maintain the turn because the required radial force is less than the maximum available friction.
• Student B adds the radial and tangential force magnitudes directly and concludes that the car cannot maintain the turn.

(a) Determine whether the car can maintain the requested motion without slipping.

(b) If the car must maintain a speed of \(12\text{ m s}^{-1}\), calculate the maximum possible tangential acceleration before slipping occurs.

(c) If the tangential acceleration must remain at \(4.0\text{ m s}^{-2}\), calculate the maximum speed at which the car can maintain the circular path without slipping.

(d) Explain the errors in both students’ reasoning.

Solution 5

The car cannot maintain the requested motion at \(12\text{ m s}^{-1}\) with a tangential acceleration of \(4.0\text{ m s}^{-2}\). The required horizontal contact force is approximately \(5.38\text{ kN}\), exceeding the maximum available \(4.90\text{ kN}\).

Student A ignores the tangential force requirement. Student B uses the wrong method to combine perpendicular force components. Although Student B reaches the correct conclusion, their reasoning is incorrect.

Step 1: Calculate the maximum available static friction.

Because the road is level and there is no downforce, the normal force balances the car’s weight.

\[
N=mg
\]

\[
N=1000\times9.8
\]

\[
N=9800\text{ N}
\]

The maximum static friction is:

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

Substitute:

\[
\begin{aligned}
F_{s,\max}
&=0.50\times9800\\
&=4900\text{ N}
\end{aligned}
\]

This is the maximum magnitude of the combined horizontal contact force.

Step 2: Calculate the required radial force.

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

\[
\begin{aligned}
F_r&=\frac{1000\times12^2}{40}\\
&=3600\text{ N}
\end{aligned}
\]

Step 3: Calculate the required tangential force.

\[
F_t=ma_t
\]

\[
F_t=1000\times4.0
\]

\[
F_t=4000\text{ N}
\]

Step 4: Combine the two components.

The radial and tangential components are perpendicular.

Therefore, the required total horizontal contact force is:

\[
F_s=\sqrt{F_r^2+F_t^2}
\]

Substitute:

\[
\begin{aligned}
F_s&=\sqrt{3600^2+4000^2}\\
&=\sqrt{12\,960\,000+16\,000\,000}\\
&=\sqrt{28\,960\,000}\\
&=5381\text{ N}
\end{aligned}
\]

Compare this with the maximum available friction:

\[
5381\text{ N}>4900\text{ N}
\]

Therefore, the requested motion is impossible without slipping under the stated assumptions.

Step 5: Find the maximum tangential acceleration at a speed of \(12\text{ m s}^{-1}\).

At this fixed speed, the radial force requirement remains \(3600\text{ N}\).

The remaining friction capacity must provide the tangential component.

Using Pythagoras:

\[
F_{t,\max}=
\sqrt{F_{s,\max}^2-F_r^2}
\]

Substitute:

\[
\begin{aligned}
F_{t,\max}
&=\sqrt{4900^2-3600^2}\\
&=\sqrt{11\,050\,000}\\
&=3324\text{ N}
\end{aligned}
\]

Using \(F_t=ma_t\):

\[
a_{t,\max}=\frac{3324}{1000}
\]

Therefore:

\[
\boxed{a_{t,\max}=3.32\text{ m s}^{-2}}
\]

At \(12\text{ m s}^{-1}\), the car cannot accelerate tangentially at \(4.0\text{ m s}^{-2}\) without exceeding the friction limit.

Step 6: Find the maximum speed when tangential acceleration is \(4.0\text{ m s}^{-2}\).

The tangential force requirement remains:

\[
F_t=4000\text{ N}
\]

The maximum radial force available is:

\[
F_{r,\max}=
\sqrt{F_{s,\max}^2-F_t^2}
\]

Substitute:

\[
\begin{aligned}
F_{r,\max}
&=\sqrt{4900^2-4000^2}\\
&=\sqrt{8\,010\,000}\\
&=2830\text{ N}
\end{aligned}
\]

Now use the radial force equation:

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

Rearrange:

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

Substitute:

\[
\begin{aligned}
v_{\max}
&=\sqrt{\frac{2830\times40}{1000}}\\
&=\sqrt{113.2}\\
&=10.64\text{ m s}^{-1}
\end{aligned}
\]

Therefore:

\[
\boxed{v_{\max}\approx10.6\text{ m s}^{-1}}
\]

Step 7: Explain the two students’ errors.

Student A compares only the radial force with maximum friction.

This ignores the fact that the car also needs a tangential force to increase its speed.

Student B adds the radial and tangential magnitudes directly.

However, these components are perpendicular, so their resultant must be calculated using Pythagoras.

The deeper point is that the tyres have one maximum available resultant force. The radial and tangential requirements must both fit within that limit.

Adding an extra centripetal-force arrow would also be incorrect because the radial component is already included in the required resultant.

This is why a correct free-body diagram must come before deciding whether the available force is sufficient.

09What to apply next

The most reliable way to approach circular-motion questions is to distinguish three things:

• The actual forces acting on the object.
• The radial and tangential components of the resultant force.
• The acceleration associated with each component.

Once you can separate these ideas, you can solve problems involving tension, friction, normal forces, vertical loops, and orbital motion without double-counting forces.

A particularly useful next application is satellite motion, where gravity is the actual physical force supplying the entire inward radial resultant.

See Gravity as Centripetal Force in Circular Orbits for HSC Physics.