Can an Object Move in a Circle with Zero Net Force?

Imagine a puck travelling around a circular air table at a steady \(3.1\text{ m s}^{-1}\). Its speed never changes. A student checks the speed, sees the same reading every second, and concludes that the acceleration must be zero.

But the puck keeps turning. Something must be changing its velocity.

Can an object move in a circle if its net force is zero? No, not in an inertial reference frame. Circular motion at a non-zero speed requires a net force directed towards the centre of the circle. The speed can remain constant, but the velocity cannot, because its direction changes continuously.

There is an interesting complication: an observer rotating alongside the object can describe it as stationary, with zero acceleration relative to their frame. Understanding why both descriptions can be correct requires separating real forces from apparent forces.

01Why constant speed does not mean zero acceleration

Start by comparing two objects travelling at \(5.0\text{ m s}^{-1}\).

• Object A travels in a straight line at constant speed.
• Object B travels around a circular track at constant speed.

Both have the same speed. Do they have the same acceleration?

No. Object A maintains the same velocity because neither its speed nor its direction changes. Object B continually changes direction, so its velocity changes even though its speed stays fixed.

This is the distinction between speed and velocity:

• Speed is a scalar. It tells you how fast an object moves.
• Velocity is a vector. It tells you both how fast an object moves and its direction of motion.

An object accelerates whenever its velocity changes. It does not need to speed up or slow down.

What happens to velocity during circular motion?

Consider a puck travelling clockwise around a circle.

At the top of its path, its velocity points horizontally to the right. A moment later, its velocity points slightly downwards and to the right. A little later, it points more steeply downwards.

The direction changes continuously.

At every instant, the velocity points along the tangent to the circular path. The tangent is perpendicular to the radius connecting the object to the centre.

Here is the important observation: even though the velocity vectors have equal magnitudes, they point in different directions.

The change in velocity, \(\Delta v\), points approximately towards the centre over a short time interval. This tells us that the acceleration is directed inwards.

This acceleration is called centripetal acceleration.

Its magnitude is:

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

where:

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

The equation tells us something important: an object travelling faster around the same circle requires greater inward acceleration.

Using Newton’s second law, the required inward resultant force is:

\[ F_{r,\mathrm{net}}=\frac{mv^2}{r} \]

Here, \(m\) is the object’s mass in kilograms and \(F_{r,\mathrm{net}}\) is the magnitude of the net radial force in newtons.

This is often called the centripetal force. It is not an additional type of physical force. It describes the inward resultant of the actual forces acting on the object.

Depending on the situation, that resultant might come from tension, friction, gravity, or a combination of forces.

Compare the two situations

The key distinction is direction.

Zero net force means zero acceleration, which means constant velocity. It does not mean an object must be stationary.

For a fuller explanation of this relationship, see Newton’s First Law: Inertia and Zero Net Force Explained.

02An experiment: measuring the force needed for circular motion

It is one thing to calculate the force required for circular motion. It is another to measure it.

A useful investigation involves a small puck moving around a horizontal air table.

Setting up the apparatus

Imagine a horizontal air table with a small central post.

A light string connects a puck to the central post. An inline force sensor measures the tension in the string. The puck is free to move around the post, with the string maintaining a fixed radius.

The air table reduces friction by supporting the puck on a thin cushion of air. Its normal force balances its weight, leaving the horizontal string tension as the main horizontal force.

A ruler measures the radius. A fixed overhead camera records the puck’s position, and a timer measures the time taken for each revolution.

Assume that:

• The string is light and remains taut.
• The puck moves at constant speed.
• Horizontal friction and air resistance are negligible.
• The force sensor is calibrated.
• The puck follows a circle of constant radius.

What should the experiment show?

Suppose a student initially uses only a timer to measure the time taken for each revolution.

The timer records a consistent period. The student concludes that the puck has constant speed and therefore zero acceleration.

There is a problem with this design.

The timer measures how quickly the puck completes each revolution. It does not directly show how the direction of its velocity changes. Constant speed alone cannot establish zero acceleration.

We need to improve the investigation.

Revising the measurement

Add two measurements:

1. A fixed overhead camera records the puck’s position at regular time intervals.
2. The inline force sensor measures the tension in the string.

The camera now reveals the changing direction of motion. The force sensor measures the inward pull maintaining the circular path.

If the model is correct, the measured tension should approximately match:

\[ F_{\mathrm{tension}}=\frac{mv^2}{r} \]

This equality assumes that the string tension is the only significant horizontal force.

Illustrative experimental data

Consider a puck with a mass of \(0.20\text{ kg}\), travelling in a circle of radius \(0.50\text{ m}\).

The following values are illustrative, not measurements from an actual experiment.

Notice what happens when the period halves.

The puck completes each revolution in half the time, so its tangential speed doubles. However, the required inward force increases by a factor of four.

This is consistent with:

\[ F_{r,\mathrm{net}}=\frac{mv^2}{r} \]

At constant mass and radius, the force is proportional to the square of the speed.

The agreement between the calculated force and the illustrative sensor readings supports the model.

The experiment also highlights an important distinction: the sensor measures string tension. We identify that tension as the centripetal force because it supplies the inward resultant in this particular arrangement.

In another experiment, friction or gravity might provide the required inward resultant instead.

03What if the object’s speed changes?

So far, we have considered uniform circular motion, where the speed remains constant.

But an object can travel around a circle while speeding up or slowing down.

In that situation, its acceleration has two perpendicular components.

Radial acceleration

The radial component points towards the centre. It changes the direction of velocity.

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

Tangential acceleration

The tangential component acts along the tangent to the circular path. It changes the magnitude of velocity, meaning the object’s speed.

For constant tangential acceleration:

\[ a_t=\frac{\Delta v}{\Delta t} \]

where \(\Delta v\) is the change in speed over the time interval \(\Delta t\).

Newton’s second law applies separately to these perpendicular components:

\[ F_r=ma_r \]

\[ F_t=ma_t \]

Because the components are perpendicular, the magnitude of the total resultant force is:

\[ F_{\mathrm{net}}=\sqrt{F_r^2+F_t^2} \]

This gives us an important result.

Even when an object speeds up or slows down while travelling around a circle, it still needs an inward radial resultant force.

A tangential force cannot replace the required radial force.

04Does the answer change in a rotating reference frame?

There is one situation that can make the original question seem confusing.

Imagine a puck attached to a rotating turntable. It stays at a fixed distance from the centre and rotates with the platform.

A camera fixed to the laboratory sees the puck moving in a circle.

A camera attached to the turntable sees the puck sitting perfectly still.

Which observer is correct?

Both are describing the same physical situation.

The difference is their reference frame.

The laboratory observer

The laboratory observer uses an approximately inertial reference frame. This is a frame that is not accelerating or rotating.

From this perspective:

• The puck travels in a circular path.
• Its velocity continually changes direction.
• Its acceleration points towards the centre.
• The real resultant force points towards the centre.

The net inward force is:

\[ F_{r,\mathrm{net}}=\frac{mv^2}{r} \]

There is no real outward force balancing this inward resultant.

The rotating observer

The observer attached to the turntable uses a rotating reference frame.

This frame is non-inertial because it is rotating and therefore accelerating.

The puck appears stationary relative to the observer. Its position and velocity relative to the platform do not change.

However, the inward real force has not disappeared.

To apply Newton’s laws in this rotating frame, the observer introduces an apparent outward force called the centrifugal force.

For an object stationary relative to a uniformly rotating frame, its magnitude is:

\[ F_{\mathrm{cf}}=m\omega^2r \]

where:

• \(F_{\mathrm{cf}}\) is the centrifugal pseudo-force, in newtons.
• \(m\) is the object’s mass, in kilograms.
• \(\omega\) is the angular speed of the rotating frame, in radians per second.
• \(r\) is the object’s distance from the rotation axis, in metres.

Since \(v=\omega r\), this can also be written as:

\[ F_{\mathrm{cf}}=\frac{mv^2}{r} \]

In the rotating frame, the outward centrifugal pseudo-force balances the inward real force.

The effective force sum in that frame is therefore zero.

Comparing the two descriptions

The important distinction is that zero relative acceleration does not necessarily mean zero real resultant force.

The rotating observer needs the centrifugal pseudo-force to make Newton’s second law consistent with their non-inertial frame.

The two observers are not disagreeing about what physically happens. They are using different reference frames.

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

05A reliable way to approach exam questions

When you encounter a circular-motion problem, work through these distinctions before choosing an equation.

A useful habit is to identify the observer’s reference frame before applying Newton’s second law.

For standard circular-motion calculations, start with the inertial laboratory frame. Identify the actual forces, resolve them into radial and tangential components, and calculate their resultant.

Do not assume that constant speed means zero net force.

06Questions and solutions

Question 1

A \(0.30\text{ kg}\) puck travels at a constant speed of \(4.0\text{ m s}^{-1}\) around a horizontal circular path of radius \(2.0\text{ m}\).

Calculate the magnitude and direction of the net force acting on the puck.

Explain why the constant speed does not mean the net force is zero.

Solution 1

The net force is \(2.4\text{ N}\) towards the centre of the circle. It is not zero because the puck’s velocity changes direction continuously.

Step 1: Calculate the radial acceleration.

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

Substituting:

\[ a_r=\frac{4.0^2}{2.0} \]

\[ a_r=8.0\text{ m s}^{-2} \]

Step 2: Calculate the net force.

Using Newton’s second law:

\[ F_{\mathrm{net}}=ma_r \]

\[ F_{\mathrm{net}}=0.30\times8.0 \]

\[ \boxed{F_{\mathrm{net}}=2.4\text{ N}} \]

The direction is towards the centre.

The puck maintains constant speed, but its direction changes. Therefore, its velocity changes, producing inward acceleration and a non-zero net force.

Question 2

An \(1100\text{ kg}\) car travels around a level circular bend of radius \(60\text{ m}\) at a constant speed of \(18\text{ m s}^{-1}\).

Assume that:

• Static friction between the tyres and road is the only horizontal force.
• The road is level.
• The normal force balances the car’s weight.
• Gravitational acceleration is \(9.8\text{ m s}^{-2}\).

Calculate:

a) The required inward net force.

b) The minimum coefficient of static friction needed to maintain the circular path.

c) Whether the car can maintain its circular path at \(18\text{ m s}^{-1}\) if the coefficient of static friction is only \(0.40\). Determine the maximum speed it could maintain instead.

Solution 2

The required inward force is \(5940\text{ N}\), and the minimum coefficient of static friction is approximately \(0.55\). With a coefficient of \(0.40\), the car cannot maintain the bend at \(18\text{ m s}^{-1}\); its maximum speed is approximately \(15.3\text{ m s}^{-1}\).

Step 1: Calculate the required inward resultant force.

The friction force must provide the centripetal force:

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

Substituting:

\[ F_r=\frac{1100\times18^2}{60} \]

\[ F_r=\frac{1100\times324}{60} \]

\[ \boxed{F_r=5940\text{ N}} \]

Step 2: Calculate the normal force.

Because the road is horizontal, the vertical forces balance.

\[ N=mg \]

\[ N=1100\times9.8 \]

\[ N=10780\text{ N} \]

Step 3: Calculate the minimum friction coefficient.

The maximum static friction is:

\[ f_{\mathrm{max}}=\mu_sN \]

To maintain the circular path:

\[ \mu_sN\geq F_r \]

Therefore:

\[ \mu_s\geq\frac{5940}{10780} \]

\[ \boxed{\mu_s\geq0.55} \]

The road must provide a coefficient of static friction of approximately 0.55 or greater.

Step 4: Calculate the maximum speed when \(\mu_s=0.40\).

At maximum speed, friction is at its limiting value:

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

Cancel the mass and rearrange:

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

Substituting:

\[ v_{\max}=\sqrt{0.40\times9.8\times60} \]

\[ v_{\max}=\sqrt{235.2} \]

\[ \boxed{v_{\max}=15.3\text{ m s}^{-1}} \]

The car cannot maintain the required circular path at \(18\text{ m s}^{-1}\) because the available friction is insufficient.

Notice that friction does not need to point forwards to maintain the car’s speed. It acts towards the centre, changing the direction of velocity.

Question 3

A \(1.5\text{ kg}\) drone follows a horizontal circular route with a radius of \(20\text{ m}\).

Its speed sensor records \(8.0\text{ m s}^{-1}\) at every measurement point.

A student argues:

“The speed never changes, so the drone has zero acceleration and zero net force.”

a) Explain the error in this argument.

b) Calculate the drone’s radial acceleration.

c) Calculate the magnitude and direction of its net force.

d) Explain what additional information a fixed overhead video would provide that a speed sensor alone cannot.

Solution 3

The student is incorrect. The drone has an inward acceleration of \(3.2\text{ m s}^{-2}\) and a net force of \(4.8\text{ N}\) towards the centre.

Step 1: Explain why constant speed does not mean zero acceleration.

The speed sensor measures only the magnitude of the drone’s velocity.

It does not measure its direction.

As the drone follows the circular route, its velocity direction changes continuously. Therefore, its velocity changes even though its speed remains constant.

This means the drone accelerates.

Step 2: Calculate the radial acceleration.

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

Substituting:

\[ a_r=\frac{8.0^2}{20} \]

\[ \boxed{a_r=3.2\text{ m s}^{-2}} \]

Step 3: Calculate the net force.

\[ F_{\mathrm{net}}=ma_r \]

\[ F_{\mathrm{net}}=1.5\times3.2 \]

\[ \boxed{F_{\mathrm{net}}=4.8\text{ N}} \]

The force acts towards the centre of the circular route.

Step 4: Explain the role of the overhead video.

The video records the drone’s positions at successive time intervals.

From these positions, we can determine the direction of its velocity at different points. We can then compare successive velocity vectors.

Even though their magnitudes are equal, their directions differ. The change in velocity reveals the inward acceleration.

The speed sensor alone cannot establish zero acceleration because it does not record the direction of velocity.

Question 4

A \(0.50\text{ kg}\) ball is attached to a light string of length \(1.0\text{ m}\) and moves in a vertical circle.

At the lowest point of its path, its speed is \(6.0\text{ m s}^{-1}\).

Assume air resistance is negligible and take gravitational acceleration as \(9.8\text{ m s}^{-2}\).

Calculate:

a) The magnitude and direction of the net radial force.

b) The tension in the string.

c) Explain why the tension is not equal to the centripetal force in this situation.

Solution 4

The net radial force is \(18.0\text{ N}\) upwards, and the string tension is \(22.9\text{ N}\) upwards. The tension is greater than the net radial force because gravity acts downwards, opposing the required inward resultant.

Step 1: Identify the inward direction.

At the lowest point of the circle, the centre is directly above the ball.

Therefore, the inward radial direction is upwards.

The string tension acts upwards, while gravity acts downwards.

Step 2: Calculate the required net radial force.

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

Substituting:

\[ F_r=\frac{0.50\times6.0^2}{1.0} \]

\[ F_r=0.50\times36 \]

\[ \boxed{F_r=18.0\text{ N upward}} \]

Step 3: Calculate the ball’s weight.

\[ F_g=mg \]

\[ F_g=0.50\times9.8 \]

\[ F_g=4.9\text{ N downward} \]

Step 4: Calculate the tension.

Take upwards, towards the centre, as positive.

Using Newton’s second law:

\[ T-F_g=F_r \]

Substituting:

\[ T-4.9=18.0 \]

\[ T=18.0+4.9 \]

\[ \boxed{T=22.9\text{ N}} \]

Step 5: Explain the difference.

The centripetal force is the net inward resultant, not necessarily the tension alone.

In this example, the tension acts inwards, but gravity acts outwards at the lowest point.

The resultant is therefore:

\[ F_{\mathrm{net}}=22.9-4.9 \]

\[ F_{\mathrm{net}}=18.0\text{ N inward} \]

The string tension and the centripetal resultant are different quantities. Confusing them is a common error in vertical circular-motion questions.

Question 5

A \(0.20\text{ kg}\) puck is attached to the centre of a motorised turntable by a light horizontal string.

The puck remains at a radius of \(0.50\text{ m}\) and completes one revolution every \(1.00\text{ s}\).

An inline force sensor records a tension of \(3.9\text{ N}\) towards the centre.

Three observers watch the same puck:

• Observer A stands in the laboratory. They see the puck travelling in a circle.
• Observer B sits on a train travelling in a straight line at a constant velocity of \(12\text{ m s}^{-1}\). The train’s axes remain parallel to the laboratory axes. They see the puck follow a path that is not circular relative to the train.
• Observer C sits on the rotating turntable. They see the puck remaining stationary relative to the platform.

A student makes the following argument:

“Observer B does not see a circular path, and Observer C does not see any movement. Therefore, the net force must depend on the observer. It could be zero for B or C even though it is non-zero for A.”

Evaluate this argument.

In your response:

a) Calculate the real resultant force required to maintain the puck’s circular motion in the laboratory frame and compare it with the sensor reading.

b) Explain whether Observer B can measure zero acceleration simply because the path appears different.

c) Explain how Observer C can describe the puck as stationary while still accounting for the inward real force.

d) Decide whether any of the three observers can claim that the resultant of the real interaction forces is zero.

Solution 5

The student’s argument is incorrect. The real resultant of the interaction forces is approximately \(3.95\text{ N}\) towards the centre in every inertial frame. Observer C can describe zero acceleration relative to the rotating platform, but must introduce an outward centrifugal pseudo-force to apply Newton’s second law in that frame.

Step 1: Calculate the puck’s tangential speed.

The period is the time taken to complete one revolution.

For circular motion:

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

Substituting:

\[ v=\frac{2\pi\times0.50}{1.00} \]

\[ \boxed{v=3.14\text{ m s}^{-1}} \]

Step 2: Calculate the radial acceleration.

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

Substituting:

\[ a_r=\frac{3.14^2}{0.50} \]

\[ \boxed{a_r=19.7\text{ m s}^{-2}} \]

The acceleration is directed towards the centre.

Step 3: Calculate the real resultant force.

Using Newton’s second law:

\[ F_{\mathrm{net}}=ma_r \]

\[ F_{\mathrm{net}}=0.20\times19.7 \]

\[ \boxed{F_{\mathrm{net}}=3.94\text{ N inward}} \]

The calculated value is approximately \(3.95\text{ N}\), consistent with the force sensor reading of \(3.9\text{ N}\), allowing for rounding and measurement uncertainty.

Observer A’s description is therefore consistent with the measurements.

Step 4: Consider Observer B’s train frame.

Observer B is travelling at constant velocity in a straight line.

This is an inertial reference frame, just like the laboratory frame.

The puck’s path relative to the train does not have to be circular. The train is moving relative to the laboratory, so the puck’s position is measured relative to a translating origin.

However, the train is not accelerating.

Adding or subtracting a constant velocity does not change acceleration.

Therefore, Observer B still measures the puck’s acceleration as:

\[ a_r=19.7\text{ m s}^{-2} \]

The real resultant force remains approximately \(3.95\text{ N}\).

A different-looking trajectory does not imply zero acceleration.

Step 5: Consider Observer C’s rotating frame.

Observer C rotates with the turntable.

The puck remains at a fixed position relative to this observer, so its relative acceleration is zero.

However, this observer is in a non-inertial frame because the turntable is rotating.

To apply Newton’s second law in this frame, Observer C must include an outward centrifugal pseudo-force.

The angular speed is:

\[ \omega=\frac{2\pi}{P} \]

\[ \omega=\frac{2\pi}{1.00} \]

\[ \omega=6.28\text{ rad s}^{-1} \]

The centrifugal pseudo-force is:

\[ F_{\mathrm{cf}}=m\omega^2r \]

Substituting:

\[ F_{\mathrm{cf}} =0.20\times6.28^2\times0.50 \]

\[ \boxed{F_{\mathrm{cf}}=3.95\text{ N outward}} \]

This balances the \(3.95\text{ N}\) inward real force in the rotating-frame equation.

The effective force sum is zero, which is consistent with the puck being stationary relative to the rotating platform.

Step 6: Evaluate the student’s conclusion.

The three observers describe the same physical situation differently:

• Observer A sees circular motion and a real inward resultant force.
• Observer B sees a different trajectory but measures the same non-zero acceleration because their frame is still inertial.
• Observer C sees the puck stationary relative to the platform and uses an outward centrifugal pseudo-force to balance the inward real force in their rotating-frame description.

The important distinction is between the resultant of real interaction forces and the effective force sum used in a non-inertial reference frame.

No inertial observer can claim that the real resultant is zero while this puck continues moving in a circle at a non-zero speed.

This is the important limitation of the phrase “zero net force”: you must know which reference frame is being used and whether apparent forces have been included.

07Where to take this next

The central idea is straightforward: constant speed does not mean constant velocity, and constant velocity is what zero net force produces in an inertial frame.

Whenever you encounter circular motion, identify the inward radial resultant before deciding whether the forces are balanced.

The next step is identifying which real interaction supplies that resultant. In a particular problem, it might be string tension, friction, gravity, or several forces acting together. That is where circular-motion free-body diagrams become essential.