Centripetal vs Centrifugal Force: The Key Differences

A puck is held at a fixed radius on a rotating turntable. A camera bolted to the laboratory bench sees it travelling in a circle. Someone rotating with the turntable sees the same puck sitting perfectly still.

So which force is acting on it? Is it being pulled towards the centre or pushed away from it?

The distinction is straightforward once you identify the observer’s frame of reference:

• Centripetal force is the net real force directed towards the centre of a circular path. It is required to produce centripetal acceleration.
• Centrifugal force is an apparent outward force introduced when describing motion from a rotating, non-inertial frame of reference. It is not a real interaction force acting on the object in an inertial frame.

They are not two real forces acting on an object simultaneously in the laboratory frame. Nor are they a Newton’s third-law action-reaction pair.

Before looking at the equations, consider one prediction. If the turntable spins twice as fast while the puck stays at the same radius, does the string tension double or quadruple?

It quadruples. Understanding why will also help you distinguish the object’s tangential speed from the force responsible for its circular motion.

01A rotating-platform investigation: where does centrifugal force come from?

A useful way to understand the difference is to investigate a rotating object from two different viewpoints.

This is a proposed model investigation using illustrative measurements, rather than a report of an actual experimental data set.

Setting up the experiment

Imagine a horizontal, low-friction turntable with a small puck resting on its surface.

A light string connects the puck to a force sensor mounted at the central hub. The puck sits a distance \(r\) from the centre. The sensor and turntable rotate together.

Three other measurements are available:

• A ruler measures the puck’s distance from the centre.
• An optical tachometer counts the turntable’s revolutions, allowing its period to be calculated.
• A stationary overhead camera records the puck’s motion relative to the laboratory.

A second camera rotates with the turntable and records the motion relative to the platform.

The force sensor transmits its reading to a stationary display.

Assume the string remains taut, the turntable rotates at constant angular speed, and horizontal friction and air resistance are negligible.

The string exerts an inward tension force on the puck. The puck exerts an equal outward force on the string and sensor.

The sensor measures the magnitude of this tension. It does not measure a centrifugal force.

What should the experiment show?

Start with the laboratory observer.

The puck travels around the centre in a circular path. Even if its speed remains constant, its velocity changes direction continuously.

Therefore, it must accelerate towards the centre.

The string supplies the inward force responsible for this acceleration.

Now consider the observer rotating with the turntable.

The puck appears stationary. Its position relative to the platform does not change.

But the string is still pulling it towards the centre. To explain why the puck remains stationary in this rotating frame using Newton’s laws, the observer introduces an outward centrifugal force.

The two descriptions are consistent:

• The laboratory observer sees circular motion caused by an inward resultant force.
• The rotating observer sees a stationary object, with the inward real force balanced by an outward centrifugal pseudo-force.

The problem with the original observation

Suppose we only use the camera attached to the turntable.

It records a stationary puck, so we might conclude that the puck has zero acceleration and therefore zero net force.

That conclusion is incomplete.

The camera itself is rotating. It is not an inertial observer, and its measurements describe motion relative to an accelerating reference frame.

A stationary appearance in a rotating frame does not mean the object has zero acceleration in the laboratory frame.

We need another observation.

Revising the experiment: release the string

Add a quick-release mechanism to the string.

Once the puck is rotating steadily, cut the string while keeping the turntable motor running.

What should the stationary camera observe?

The puck travels in a straight line tangent to its original circular path. With the string removed and friction negligible, there is no horizontal real force to bend its trajectory.

The rotating camera records something different.

Relative to the turntable, the puck appears to move outwards and fall behind the rotating platform.

The difference comes from the reference frames, not from two different physical events.

The stationary observer sees the puck continue along its existing velocity. The rotating observer sees the puck move relative to a platform that continues turning beneath it.

The revised experiment addresses the original flaw: we can now compare the trajectory against a stationary reference rather than relying entirely on observations made from the rotating platform.

The important conclusion: the string tension is a real inward force. Centrifugal force is an additional term used to describe motion in the rotating frame. The force sensor measures the former, not the latter.

01Understand the direction of motion before considering forces

The most common mistake in circular motion is to assume that an object moves in the direction of the force acting on it.

That is not generally true.

An object’s velocity describes the direction in which it is moving. Its acceleration describes how its velocity is changing.

For an object travelling around a circle, its instantaneous velocity is tangent to the circular path.

Its centripetal acceleration points towards the centre.

Imagine swinging a ball around your head on a string.

At every instant, the ball’s velocity points along the tangent to its circular path. The string pulls towards your hand, changing the direction of the ball’s velocity.

If the ball travels at constant speed, the magnitude of its velocity stays the same, but its direction changes.

That means it still accelerates.

This is the key to understanding centripetal force.

Calculating centripetal force

For an object of mass \(m\), travelling at tangential speed \(v\) in a circle of radius \(r\), the magnitude of its centripetal acceleration is

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

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

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

where:

• \(F_c\) is the magnitude of the centripetal net force, measured in newtons (N).
• \(m\) is the object’s mass, measured in kilograms (kg).
• \(v\) is its tangential speed, measured in metres per second (m/s).
• \(r\) is the radius of the circular path, measured in metres (m).

Centripetal force is not a separate type of physical force.

It describes the inward radial component of the net real force acting on an object.

Depending on the situation, that force could be supplied by tension, friction, gravity, or a combination of forces.

For example, gravity supplies the centripetal force for a satellite in a circular orbit. The satellite does not need an additional force called centripetal force.

For a fuller explanation of identifying the actual force responsible for circular motion, see Centripetal Force: How to Identify the Real Force in HSC Physics.

What happens if the object’s speed changes?

So far, we’ve considered uniform circular motion, where speed remains constant.

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

In that situation, its acceleration has two components:

• Radial acceleration, directed towards the centre, which changes the direction of velocity.
• Tangential acceleration, directed along or opposite to the direction of motion, which changes the speed.

The tangential acceleration is

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

for constant tangential acceleration.

The corresponding tangential resultant force is

\[ F_t=ma_t \]

The radial and tangential components are perpendicular.

This means that when an object speeds up while moving around a circle, its total acceleration is not purely towards the centre.

That distinction becomes important in the second worked example.

02What exactly is centrifugal force?

Centrifugal force appears when you describe motion from a rotating reference frame.

A rotating reference frame is non-inertial because it is accelerating relative to an inertial frame.

Newton’s laws in their usual form apply directly in inertial frames. To use Newton’s second law in a rotating frame, we introduce additional apparent forces, also called pseudo-forces or inertial forces.

One of these is centrifugal force.

For an object at a fixed position relative to a frame rotating at constant angular speed, its magnitude is

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

where:

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

The centrifugal force acts radially outwards.

Angular speed is related to tangential speed by

\[ v=\omega r \]

Therefore,

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

Notice something important.

For an object stationary relative to a uniformly rotating platform, the magnitude of the centrifugal force equals the magnitude of the centripetal resultant force.

However, the two terms belong to different reference-frame descriptions.

Comparing the two reference frames

Consider the puck attached to the centre of a rotating turntable.

In the laboratory frame, the real forces have an inward resultant.

In the rotating frame, the puck is stationary relative to the platform. Its real inward force is balanced by the outward centrifugal pseudo-force in the rotating-frame equation.

The distinction is not that one observer is correct and the other is wrong.

They are describing the same physical situation using different reference frames.

Is centrifugal force a Newton’s third-law reaction force?

No.

This is a particularly tempting misconception because centripetal and centrifugal forces have equal magnitudes and opposite directions in the stationary-object rotating-frame example.

But Newton’s third law describes interactions between two objects.

Consider the string again:

• The string pulls the puck towards the centre.
• The puck pulls the string away from the centre.

These are an action-reaction pair. They act on different objects.

Centrifugal force is not the reaction to the string tension. It is a pseudo-force introduced because the observer is using a rotating reference frame.

Equal magnitude and opposite direction do not automatically make two forces a Newton’s third-law pair.

03A familiar example: why you feel pushed outwards in a turning car

Imagine sitting in the passenger seat of a car travelling around a left-hand bend at constant speed.

You feel yourself pressing against the right-hand door.

It seems as though something is pushing you outwards.

From the road’s reference frame, there is no real outward force on you.

Your body tends to continue in a straight line because of inertia. The car changes direction, and the seat or door exerts an inward force on your body to make it follow the bend.

You also exert an outward force on the door.

From the car’s non-inertial frame, you can introduce an apparent centrifugal force acting outwards. It balances the inward contact force while you remain stationary relative to the car.

The sensation is real. The interpretation of an outward force depends on the reference frame.

This is the same distinction as the rotating puck, just with a much more familiar object.

04Worked examples

05A quick way to decide which force to use

When a circular-motion question mentions centripetal or centrifugal force, identify the reference frame before writing an equation.

A reliable HSC problem-solving sequence is:

1. Identify the observer’s reference frame.
2. Draw the object’s circular path and mark its instantaneous velocity.
3. Identify the real forces acting on the object.
4. Resolve the real forces into radial and tangential components.
5. Apply \(F_r=mv^2/r\) for the inward radial resultant in an inertial frame.
6. Only introduce centrifugal force if the question uses a rotating reference frame.

One final qualification: if an object moves relative to a rotating platform, centrifugal force alone is not enough to describe its motion in that frame. Additional apparent effects, including the Coriolis force, may be required. You generally won’t need to calculate these for the standard HSC circular-motion problems, but recognising the limitation prevents an incorrect explanation.

The same distinction is useful when studying orbital motion. Gravity supplies the actual inward centripetal force on a satellite, while an apparent centrifugal force can be introduced when describing the orbit from a rotating reference frame. See Gravity as Centripetal Force in Circular Orbits for HSC Physics.

07Questions and solutions

Question 1

A puck of mass \(0.40\text{ kg}\) moves at a constant speed of \(5.0\text{ m/s}\) in a horizontal circle of radius \(2.0\text{ m}\).

A light string provides the only horizontal force acting on the puck.

Calculate the centripetal net force and state its direction. Would a stationary laboratory observer include a centrifugal force in the free-body diagram?

Solution 1

The centripetal net force is \(5.0\text{ N}\) towards the centre. The laboratory observer does not include a centrifugal force.

Step 1: Apply the centripetal force equation.

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

Step 2: Substitute the values.

\[ F_c=\frac{0.40\times5.0^2}{2.0} \]

\[ F_c=\frac{10}{2} \]

\[ \boxed{F_c=5.0\text{ N inward}} \]

The string provides the inward tension responsible for changing the puck’s direction of motion.

The puck’s speed is constant, but its velocity changes direction continuously. That is why it requires an inward resultant force.

The centrifugal force would only be introduced if we described the puck from a rotating reference frame.

Question 2

A \(0.30\text{ kg}\) puck is connected to the centre of a rotating horizontal platform by a light radial string.

The puck remains at a radius of \(0.45\text{ m}\). The platform completes one revolution every \(1.20\text{ s}\).

Calculate:

1. The puck’s tangential speed.
2. The magnitude of the centripetal net force.
3. The centrifugal force in the rotating platform’s reference frame.

The platform then slows down so that its period doubles to \(2.40\text{ s}\).

Determine the new magnitudes of the centripetal and centrifugal forces.

Solution 2

Initially, the puck travels at \(2.36\text{ m/s}\), requiring an inward resultant force of approximately \(3.7\text{ N}\). The rotating-frame centrifugal force has the same magnitude but acts outwards. When the period doubles, both magnitudes decrease to approximately \(0.93\text{ N}\).

Step 1: Calculate the angular speed.

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

\[ \omega=\frac{2\pi}{1.20} \]

\[ \omega=5.24\text{ rad/s} \]

Step 2: Calculate the tangential speed.

\[ v=\omega r \]

\[ v=5.24\times0.45 \]

\[ \boxed{v=2.36\text{ m/s}} \]

Step 3: Calculate the centripetal net force.

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

\[ F_c=\frac{0.30\times2.36^2}{0.45} \]

\[ \boxed{F_c=3.7\text{ N inward}} \]

Step 4: Determine the centrifugal force.

Because the puck is stationary relative to the rotating platform, the centrifugal pseudo-force balances the inward real force.

Therefore,

\[ \boxed{F_{\mathrm{cf}}=3.7\text{ N outward}} \]

Step 5: Consider what happens when the period doubles.

The new period is

\[ T'=2.40\text{ s} \]

Angular speed is inversely proportional to period:

\[ \omega'=\frac{\omega}{2} \]

Centripetal force depends on the square of angular speed:

\[ F_c=m\omega^2r \]

Therefore,

\[ F_c'=\frac{F_c}{4} \]

\[ F_c'=\frac{3.7}{4} \]

\[ \boxed{F_c'=0.93\text{ N inward}} \]

The centrifugal force in the rotating frame also becomes

\[ \boxed{F_{\mathrm{cf}}'=0.93\text{ N outward}} \]

The important relationship is that halving angular speed reduces the required centripetal force to one-quarter of its original value.

This is why doubling the period does not simply halve the force.

Question 3

A small puck is travelling at constant speed around a horizontal circular track. It is attached to the centre by a light string.

At the instant the puck reaches the rightmost point of the circle, the string is cut.

Assume air resistance and friction are negligible.

Describe the puck’s subsequent motion from:

1. A stationary laboratory observer’s perspective.
2. An observer rotating with the platform at the same angular speed as the puck immediately before release.

What real horizontal force acts on the puck immediately after the string is cut?

Solution 3

The stationary observer sees the puck travel in a straight line tangent to its original circular path. The rotating observer sees it move outwards and fall behind relative to the platform. Immediately after release, there is no real horizontal force acting on the puck.

Step 1: Consider the laboratory observer.

Before the string is cut, the puck has an instantaneous tangential velocity.

The string provides the inward centripetal force, continually changing the direction of this velocity.

When the string is cut, the tension disappears.

With no other horizontal force, the puck continues at constant velocity along the tangent to the circle.

Step 2: Consider the rotating observer.

The platform continues rotating after the string is cut.

The puck no longer follows the platform’s circular motion. Relative to the platform’s markings, it appears to move outwards and backwards.

The rotating observer describes this motion using inertial pseudo-forces, including centrifugal and, once relative motion develops, Coriolis effects.

Step 3: Identify the real forces.

Immediately after the string is cut:

• The string tension is zero.
• The vertical gravitational force is balanced by the normal force from the surface.
• There is no horizontal real force, under the stated assumptions.

The puck does not need an outward real force to continue moving.

Its motion follows from its existing tangential velocity and inertia.

Question 4

A car of mass \(1000\text{ kg}\) travels around a flat circular bend of radius \(50\text{ m}\).

Its speed increases uniformly from \(10\text{ m/s}\) to \(16\text{ m/s}\) over \(3.0\text{ s}\).

At the end of this interval, calculate:

1. The radial component of the net force.
2. The tangential component of the net force.
3. The magnitude of the total net force.

A student claims that the car must experience an outward centrifugal force in the road’s reference frame because its speed is increasing.

Explain why this is incorrect.

Solution 4

The radial net force is \(5120\text{ N}\) inward, the tangential net force is \(2000\text{ N}\) forwards, and the total net force is approximately \(5500\text{ N}\). There is no centrifugal real force in the road’s inertial reference frame.

Step 1: Calculate the radial acceleration at the final speed.

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

\[ a_r=\frac{16^2}{50} \]

\[ a_r=5.12\text{ m/s}^2 \]

The acceleration acts towards the centre.

Step 2: Calculate the radial force component.

\[ F_r=ma_r \]

\[ F_r=1000\times5.12 \]

\[ \boxed{F_r=5120\text{ N inward}} \]

Step 3: Calculate the tangential acceleration.

\[ a_t=\frac{v_f-v_i}{\Delta t} \]

\[ a_t=\frac{16-10}{3.0} \]

\[ a_t=2.0\text{ m/s}^2 \]

Step 4: Calculate the tangential force.

\[ F_t=ma_t \]

\[ F_t=1000\times2.0 \]

\[ \boxed{F_t=2000\text{ N forward}} \]

Step 5: Calculate the total net force.

The radial and tangential components are perpendicular.

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

\[ F_{\mathrm{net}} =\sqrt{5120^2+2000^2} \]

\[ \boxed{F_{\mathrm{net}}\approx5500\text{ N}} \]

Step 6: Evaluate the student’s claim.

The student is confusing two different effects.

The car’s increasing speed produces a forward tangential acceleration. Its circular trajectory still requires an inward radial acceleration.

Both are supplied by the real horizontal resultant force from the road’s interaction with the tyres.

In the road’s reference frame, there is no centrifugal force acting on the car.

If the problem were described using a rotating reference frame, an outward centrifugal pseudo-force could be introduced. That is a different description, not an additional real force.

Question 5

A puck of mass \(0.25\text{ kg}\) rests on a low-friction horizontal turntable.

It is connected to the central hub by a light radial string, with a force sensor measuring the string tension.

The puck is at a radius of \(0.40\text{ m}\), and the turntable rotates anticlockwise with a constant period of \(1.00\text{ s}\).

Initially, the puck is stationary relative to the turntable.

The string is then cut, and the motor continues rotating the platform at the same angular speed.

Assume negligible friction and air resistance.

Answer the following:

(a) Calculate the puck’s tangential speed and the string tension immediately before the string is cut.

(b) An observer rotating with the platform says that an outward centrifugal force balances the inward string tension. A student argues that these two forces must form a Newton’s third-law pair. Explain why the student is wrong.

(c) Describe the puck’s motion immediately after release from the laboratory observer’s perspective. What happens to the force sensor reading?

(d) Does the centrifugal term disappear from the rotating-frame description when the string is cut? Explain why it would be incorrect to predict the puck’s entire subsequent path relative to the platform using centrifugal force alone.

For part (d), you may use the following additional information: when an object begins moving relative to a rotating reference frame, a Coriolis pseudo-force also appears. It acts sideways relative to the object’s motion in that frame. No Coriolis-force calculation is required.

Solution 5

The puck’s tangential speed is approximately \(2.51\text{ m/s}\), and the string tension is approximately \(3.95\text{ N}\) inward. Centrifugal force is not the Newton’s third-law partner of tension. After release, the laboratory observer sees straight-line tangential motion, while the rotating-frame description retains centrifugal force and requires an additional Coriolis term.

Step 1: Calculate the angular speed.

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

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

\[ \omega=6.28\text{ rad/s} \]

Step 2: Calculate the tangential speed.

\[ v=\omega r \]

\[ v=6.28\times0.40 \]

\[ \boxed{v=2.51\text{ m/s}} \]

Step 3: Calculate the inward string tension.

The string provides the entire horizontal centripetal resultant.

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

\[ F_c=\frac{0.25\times2.51^2}{0.40} \]

\[ \boxed{F_{\mathrm{tension}}=3.95\text{ N inward}} \]

Step 4: Examine the Newton’s third-law argument.

The rotating observer introduces an outward centrifugal pseudo-force of magnitude \(3.95\text{ N}\).

It balances the inward tension in the rotating-frame description because the puck is stationary relative to the platform.

However, these are not a Newton’s third-law pair.

The actual action-reaction pair is:

• The string pulls the puck inwards.
• The puck pulls the string outwards.

The two forces act on different objects and arise from the same physical interaction.

Centrifugal force is not an interaction between the puck and another object. It is introduced because the reference frame is rotating.

Step 5: Describe what happens in the laboratory frame after release.

The string is cut, so its tension immediately becomes zero.

The force sensor therefore reads zero, assuming no other force acts through its connection.

The puck retains its instantaneous tangential velocity of \(2.51\text{ m/s}\).

With negligible horizontal friction, it continues in a straight line tangent to the original circular path.

Step 6: Consider the rotating-frame description.

This is the subtle part.

The centrifugal term does not disappear when the string is cut.

It depends on the rotation of the reference frame and the object’s position relative to the rotation axis, not on whether the string is attached.

Immediately after release, the puck has zero velocity relative to the rotating platform. Therefore, the Coriolis term is initially zero.

The centrifugal term is now unbalanced, producing an initial outward relative acceleration in the rotating frame.

However, as the puck begins moving relative to the platform, the Coriolis pseudo-force also becomes relevant.

It acts sideways to the puck’s relative motion, causing its path to trail behind the platform’s rotation.

The tempting answer is that the puck must travel straight outwards because centrifugal force remains after the string is cut.

The problem is that this considers only one term in the rotating-frame description. Once the puck moves relative to the platform, the Coriolis effect must also be considered.

This is an extension of the standard HSC calculation, but it reveals the deeper principle: a force description must remain consistent with the reference frame used to describe the motion.

The most useful habit to take into other circular-motion questions is to identify the observer first. Once you know whether you’re working in an inertial or rotating frame, the distinction between centripetal and centrifugal force becomes much easier to manage.