How to Compare Two Circular Motions with the Same Period

Two objects each complete one revolution every 2.0 seconds. One moves in a circle of radius 0.20 m, while the other moves in a circle of radius 0.60 m.

They have the same period. Does that mean they have the same speed? Do they need the same inward net force?

No. For uniform circular motion measured in the same inertial frame, equal periods mean equal angular speeds. However, the object travelling around the larger circle has a greater tangential speed and greater centripetal acceleration.

The radial net force depends on both mass and radius. If the masses are equal, the object with the larger radius needs the greater inward force. If the masses differ, you must compare the product of mass and radius.

This gives us three relationships to remember:

• Same period means the same angular speed (assuming the same direction of rotation).
• Tangential speed is proportional to radius.
• Radial net force is proportional to mass multiplied by radius.

Let’s work through why these relationships hold, how to compare two objects quantitatively, and what changes when we describe the motion from a rotating reference frame.

01What does having the same period actually mean?

The period, \(T\), is the time taken to complete one revolution.

During one complete revolution, an object travels through an angular displacement of \(2\pi\) radians.

Angular speed describes how quickly this angle changes. It is calculated using:

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

where:

• \(\omega\) is angular speed, measured in \(\mathrm{rad\,s^{-1}}\).
• \(T\) is period, measured in seconds (s).

If two objects have the same period, they have the same angular speed magnitude.

For example, suppose both objects complete one revolution every 2.0 seconds.

\[ \omega=\frac{2\pi}{2.0} =\pi\ \mathrm{rad\,s^{-1}} \]

Both objects sweep through the same angle in the same time.

However, they do not necessarily travel the same distance.

Imagine two riders sitting on a rotating carousel. One sits close to the central pole, while the other sits near the outside edge.

Both complete each revolution together. But the outside rider travels around a much larger circle.

The outside rider must therefore cover more distance in the same time.

This is why equal periods do not mean equal tangential speeds.

Connecting angular speed to tangential speed

The distance travelled during one complete revolution is the circumference:

\[ d=2\pi r \]

Speed is distance divided by time, so:

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

Because \(\omega=2\pi/T\), we can also write:

\[ \boxed{v=\omega r} \]

Here, \(v\) is tangential speed in metres per second, and \(r\) is the radius in metres.

When two objects have the same period, their angular speeds are equal. Therefore:

\[ \boxed{\frac{v_B}{v_A}=\frac{r_B}{r_A}} \]

This is the first important comparison rule.

If object B moves around a circle with twice the radius of object A, B has twice the tangential speed, provided both motions are uniform and have the same period.

For a fuller explanation of how these quantities connect, see Speed, Radius and Period in Uniform Circular Motion.

02Comparing radial acceleration and net force

Now we need to consider something more interesting.

Two objects might have the same angular speed but different tangential speeds. Does that also mean they have different accelerations?

Yes.

In uniform circular motion, an object continuously changes the direction of its velocity. Even if its speed stays constant, its velocity changes, so it accelerates.

This acceleration points towards the centre of the circular path. It is called centripetal acceleration, or radial acceleration.

Its magnitude is:

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

Substituting \(v=\omega r\):

\[ a_r=\frac{(\omega r)^2}{r} \]

Therefore:

\[ \boxed{a_r=\omega^2r} \]

We can also express this in terms of period:

\[ \boxed{a_r=\frac{4\pi^2r}{T^2}} \]

Notice what this tells us.

If two objects have the same period, they have the same angular speed. Their radial accelerations are therefore proportional to their radii:

\[ \boxed{\frac{a_{r,B}}{a_{r,A}} =\frac{r_B}{r_A}} \]

The object travelling around the larger circle has the greater radial acceleration.

But what about the radial net force?

Newton’s second law states:

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

Applying this to the radial acceleration gives:

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

Substituting the expression for radial acceleration:

\[ \boxed{F_{\mathrm{net},r}=m\omega^2r} \]

Or, using the period:

\[ \boxed{F_{\mathrm{net},r}=\frac{4\pi^2mr}{T^2}} \]

This is the key equation for comparing the radial net forces of two objects with the same period.

Since their periods are equal:

\[ \boxed{ \frac{F_{\mathrm{net},r,B}} {F_{\mathrm{net},r,A}} = \frac{m_Br_B}{m_Ar_A} } \]

You must consider both mass and radius.

There is an important distinction here.

The radial net force is not an additional force that mysteriously appears whenever something moves in a circle. It is the resultant of the actual forces acting towards and away from the centre.

Depending on the situation, this resultant might be provided by tension, friction, gravity, a magnetic force, or a combination of forces.

Radial force versus tangential force

It is worth separating two ideas that students sometimes combine.

The radial component of the net force changes the direction of velocity. It is responsible for keeping the object on its circular path.

The tangential component changes the magnitude of velocity. It makes the object speed up or slow down.

For uniform circular motion, the tangential component of the net force is zero. The resultant force points entirely towards the centre.

If an object is speeding up while travelling around a circle, it has both radial and tangential acceleration.

This distinction becomes particularly important when comparing objects that have the same period but are not moving at constant speed.

03How could we test this experimentally?

Suppose you wanted to investigate whether two objects rotating with the same period really do require different inward forces.

A useful arrangement is a motorised horizontal turntable.

The question is straightforward:

If two masses rotate together at different radii, will the outer mass require a greater inward force?

Setting up the investigation

Start with a motorised turntable rotating around a vertical central spindle.

Attach two radial arms to the turntable. Place a small mass on each arm, at a different distance from the centre.

Between each mass and the central hub, install a calibrated force sensor aligned with the radial arm. The sensors measure the inward forces exerted on the masses.

An encoder attached to the central spindle records the number of revolutions. Its signal is sent to a data logger, allowing the period to be calculated accurately.

The masses should be secured at fixed radii. Assume that friction and other horizontal forces are negligible, and that the vertical normal force balances each object’s weight.

What should the experiment show?

The two masses are attached to the same rigid turntable.

This is important because it guarantees that they complete each revolution together. They have the same period, even though they travel along different circular paths.

Our prediction is:

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

If both masses are equal, the outer mass should require a greater inward force in direct proportion to its radius.

If the masses differ, the force ratio should follow the product \(mr\).

Why would two separate turntables be a weaker design?

An initial design might use two independent motorised turntables, with each motor adjusted to the same speed setting.

The problem is that matching the dial settings does not guarantee identical rotation periods. Different motor loads and small variations in rotation rate could produce different results.

A shared turntable solves this problem. Both objects are mechanically constrained to have the same angular speed.

The encoder then checks that the rotation rate remains steady.

This is an example of controlling variables in an investigation: we want differences in measured force to arise from differences in mass and radius, rather than differences in rotation rate.

An illustrative set of results

Consider the following calculated values for a proposed investigation.

The numbers below are model predictions, not reported experimental measurements.

Both masses rotate with a period of \(2.00\text{ s}\).

For this arrangement:

\[ \omega=\frac{2\pi}{2.00} =\pi\ \mathrm{rad\,s^{-1}} \]

The predicted inward force on A is:

\[ F_A=(0.20)(\pi^2)(0.15) \]

\[ F_A=0.296\text{ N} \]

For B:

\[ F_B=(0.40)(\pi^2)(0.30) \]

\[ F_B=1.18\text{ N} \]

Therefore:

\[ \frac{F_B}{F_A}=4.0 \]

The predicted force on B is four times greater.

Why?

Its mass is twice as large, and its radius is also twice as large.

Both changes contribute to the increased radial net force.

In a real investigation, sensor calibration, alignment, friction, and small variations in rotation rate would need to be considered before comparing measured values with these predictions.

The experiment also highlights an important distinction: the force sensor measures a real interaction between the radial arm and the mass. It does not directly measure a centrifugal force.

04A contrasting case: Can two objects have the same radial net force?

So far, our examples have shown objects with different radial net forces.

But here is a situation that catches students out.

Imagine two objects with the same period, where one has twice the radius but half the mass of the other.

The outer object has:

• Twice the tangential speed.
• Twice the radial acceleration.
• Half the mass.

What happens to the radial net force?

The two effects involving mass and radius exactly cancel.

05Does the reference frame change the comparison?

There is one more complication.

So far, we have described circular motion from the perspective of a stationary observer, such as someone standing next to a rotating platform.

This is approximately an inertial reference frame for ordinary HSC calculations.

But imagine sitting on the rotating platform itself.

The objects attached to the platform no longer appear to move in circles. They appear stationary.

Has their motion somehow disappeared?

No. We have changed the reference frame.

The stationary observer’s description

From the stationary laboratory frame:

• The object has tangential speed \(v=\omega r\).
• Its velocity direction changes continuously.
• It has inward radial acceleration \(a_r=\omega^2r\).
• Its radial net force is directed towards the centre.

The inward net force is a real resultant of the forces acting on the object.

There is no additional outward force acting on the object in this frame.

The rotating observer’s description

Now consider an observer rotating with the platform at the same angular speed.

An object fixed to the platform is stationary relative to this observer.

Therefore, in the rotating frame:

\[ v'=0 \]

and

\[ a'=0 \]

The object appears to have no acceleration relative to the platform.

However, the real inward force from the radial arm is still present.

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

Its magnitude is:

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

This outward pseudo-force balances the real inward force when the object is stationary relative to the steadily rotating platform.

The forces balance in the rotating frame:

\[ F_{\mathrm{inward}}=F_{\mathrm{cf}} \]

This does not mean that an actual outward interaction has appeared. The centrifugal force is introduced because the reference frame itself is rotating.

A useful analogy is sitting in a car as it accelerates forwards. You may feel as though something is pushing you backwards, even though there is no new backward interaction acting on you. Describing the motion from the accelerating car’s frame requires an apparent force.

The analogy is not exact: a rotating frame continually changes its direction of motion, so its apparent centrifugal force points radially away from the rotation axis.

Does the period stay the same in both frames?

Not necessarily.

This is a subtle point.

A stationary observer sees an object fixed to a rotating platform complete one revolution every \(T\) seconds.

An observer rotating with that platform sees the object remain stationary.

The object therefore does not complete revolutions relative to the rotating observer. Its period of circular motion relative to that observer is not \(T\).

Period, like velocity, must be understood relative to a specified reference frame.

For the standard HSC comparison of two circular motions, use the stationary inertial frame unless the question explicitly asks you to analyse a rotating frame.

For a more detailed explanation of the difference between these descriptions, see Centripetal vs Centrifugal Force: The Key Differences.

06A reliable method for comparing two circular motions

When an HSC question gives you two objects with the same period, use the following process.

First: Check the motion.

Are both objects undergoing uniform circular motion?

If they are, their angular speed remains constant.

If they are moving non-uniformly, the period alone does not tell you their instantaneous speeds.

Second: Identify the reference frame.

Are you comparing motion relative to a stationary observer or someone rotating with the system?

Do not mix measurements from different frames.

Third: Find angular speed.

For uniform circular motion:

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

Equal periods give equal angular speed magnitudes.

Fourth: Compare tangential speeds.

Use:

\[ v=\omega r \]

For equal periods, speed is proportional to radius.

Fifth: Compare radial accelerations.

Use:

\[ a_r=\omega^2r \]

For equal periods, radial acceleration is also proportional to radius.

Finally: Compare radial net forces.

Use:

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

For equal periods, compare the products \(mr\).

Remember that this gives the required radial component of the resultant force. You must still identify the actual forces responsible for producing it.

07An important limitation: Same period does not always mean the same instantaneous angular speed

Everything above has assumed uniform circular motion when using the proportional relationships.

But what if an object speeds up and slows down as it travels around its circle?

It could still complete every revolution in exactly the same period.

For example, imagine two independently driven rotating objects. Both complete one revolution in 4.0 seconds.

Object A spends more time moving slowly and then speeds up through part of its path.

Object B moves at a different rate throughout its revolution.

Their average angular speeds are equal:

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

But their instantaneous angular speeds do not have to be equal.

At any particular instant:

\[ v=r\omega \]

and:

\[ a_r=r\omega^2 \]

where \(\omega\) is the instantaneous angular speed.

Therefore, two objects with the same period can have different instantaneous tangential speeds and radial accelerations.

The equation:

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

still gives the average speed over one complete revolution, but it does not necessarily give the instantaneous speed.

If the angular speed changes, there can also be tangential acceleration:

\[ a_t=r\alpha \]

where \(\alpha\) is angular acceleration, measured in \(\mathrm{rad\,s^{-2}}\).

For most standard HSC uniform circular motion questions, you can use the simpler equal-angular-speed relationships. Just check that the question actually establishes uniform motion.

08Questions and solutions

Question 1

Two objects, A and B, move in uniform circular motion in the same stationary reference frame.

Both complete one revolution every \(3.0\text{ s}\).

Object A follows a circular path with radius \(0.30\text{ m}\), while object B follows a circular path with radius \(0.75\text{ m}\).

Calculate:

a. The tangential speed of each object.

b. The radial acceleration of each object.

c. The ratio of B’s radial acceleration to A’s radial acceleration.

Solution 1

Object B has a tangential speed of \(1.57\text{ m s}^{-1}\) and a radial acceleration of \(3.29\text{ m s}^{-2}\). Both values are 2.5 times those of object A.

Part a: Tangential speeds

First calculate the common angular speed:

\[ \omega=\frac{2\pi}{3.0} =2.09\ \mathrm{rad\,s^{-1}} \]

For object A:

\[ v_A=\omega r_A \]

\[ v_A=(2.09)(0.30) \]

\[ \boxed{v_A=0.628\ \mathrm{m\,s^{-1}}} \]

For object B:

\[ v_B=(2.09)(0.75) \]

\[ \boxed{v_B=1.57\ \mathrm{m\,s^{-1}}} \]

Part b: Radial accelerations

Using:

\[ a_r=\omega^2r \]

For A:

\[ a_{r,A}=(2.09)^2(0.30) \]

\[ \boxed{a_{r,A}=1.32\ \mathrm{m\,s^{-2}}} \]

For B:

\[ a_{r,B}=(2.09)^2(0.75) \]

\[ \boxed{a_{r,B}=3.29\ \mathrm{m\,s^{-2}}} \]

Part c: Acceleration ratio

\[ \frac{a_{r,B}}{a_{r,A}} = \frac{3.29}{1.32} \]

\[ \boxed{\frac{a_{r,B}}{a_{r,A}}=2.5} \]

This makes sense because B has 2.5 times the radius of A.

The common mistake is to assume that equal periods mean equal speeds and accelerations. They actually mean equal angular speeds. The larger circle requires more distance to be covered during each revolution.

Question 2

Two objects move in uniform circular motion with the same period of \(2.5\text{ s}\).

Their properties are:

Assume the only horizontal radial force on each object is the inward force supplied by its connecting arm.

Calculate:

a. The tangential speed of each object.

b. The radial net force on each object.

c. The ratio of the radial net force on B to that on A.

Explain why the force ratio is different from the speed ratio.

Solution 2

Object B has twice the tangential speed of A, but its radial net force is three times greater.

Part a: Tangential speeds

Calculate the common angular speed:

\[ \omega=\frac{2\pi}{2.5} =2.51\ \mathrm{rad\,s^{-1}} \]

For A:

\[ v_A=(2.51)(0.25) \]

\[ \boxed{v_A=0.628\ \mathrm{m\,s^{-1}}} \]

For B:

\[ v_B=(2.51)(0.50) \]

\[ \boxed{v_B=1.26\ \mathrm{m\,s^{-1}}} \]

Part b: Radial net forces

Use:

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

For A:

\[ F_A=(0.40)(2.51)^2(0.25) \]

\[ \boxed{F_A=0.632\text{ N}} \]

For B:

\[ F_B=(0.60)(2.51)^2(0.50) \]

\[ \boxed{F_B=1.89\text{ N}} \]

Both forces are directed towards their respective centres.

Part c: Force ratio

\[ \frac{F_B}{F_A} = \frac{1.89}{0.632} \]

\[ \boxed{\frac{F_B}{F_A}=3.0} \]

We can also obtain this directly from:

\[ \frac{F_B}{F_A} = \frac{m_Br_B}{m_Ar_A} \]

\[ = \frac{(0.60)(0.50)} {(0.40)(0.25)} =3.0 \]

The speed ratio is only 2.0 because speed depends on radius.

The force ratio is 3.0 because radial net force depends on both mass and radius.

The tempting mistake is to assume that twice the speed means twice the force. The relationship \(F=ma\) reminds us that mass also matters.

Question 3

Two objects, A and B, move in non-uniform circular motion. Each completes one revolution every \(4.0\text{ s}\).

They are independently driven, rather than attached to the same rigid rotating platform.

Their properties at particular instants are:

Calculate:

a. The instantaneous tangential speed of each object.

b. The instantaneous radial acceleration of each object.

c. The instantaneous radial net force on each object.

d. Explain why having the same period does not mean the two objects have the same instantaneous angular speed.

Can their tangential accelerations also be determined from the information supplied?

Solution 3

Object A has an instantaneous speed of \(0.80\text{ m s}^{-1}\), radial acceleration of \(1.60\text{ m s}^{-2}\), and radial net force of \(0.80\text{ N}\). Object B has corresponding values of \(0.60\text{ m s}^{-1}\), \(0.60\text{ m s}^{-2}\), and \(0.30\text{ N}\). Their tangential accelerations cannot be determined from the supplied information.

Part a: Instantaneous tangential speeds

For circular motion:

\[ v=r\omega \]

For A:

\[ v_A=(0.40)(2.0) \]

\[ \boxed{v_A=0.80\ \mathrm{m\,s^{-1}}} \]

For B:

\[ v_B=(0.60)(1.0) \]

\[ \boxed{v_B=0.60\ \mathrm{m\,s^{-1}}} \]

Part b: Instantaneous radial accelerations

Using:

\[ a_r=\omega^2r \]

For A:

\[ a_{r,A}=(2.0)^2(0.40) \]

\[ \boxed{a_{r,A}=1.60\ \mathrm{m\,s^{-2}}} \]

For B:

\[ a_{r,B}=(1.0)^2(0.60) \]

\[ \boxed{a_{r,B}=0.60\ \mathrm{m\,s^{-2}}} \]

Part c: Instantaneous radial net forces

Using:

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

For A:

\[ F_A=(0.50)(1.60) \]

\[ \boxed{F_A=0.80\text{ N}} \]

For B:

\[ F_B=(0.50)(0.60) \]

\[ \boxed{F_B=0.30\text{ N}} \]

Part d: Why are the angular speeds different?

The average angular speed of each object over one complete revolution is:

\[ \overline{\omega}=\frac{2\pi}{4.0} \]

\[ \overline{\omega}=1.57\ \mathrm{rad\,s^{-1}} \]

However, neither object moves at a constant angular speed.

They speed up and slow down during their respective revolutions. Their instantaneous angular speeds can therefore differ even though their total revolution times are identical.

The equation \(2\pi/T\) gives their average angular speed, not necessarily their instantaneous angular speed.

Tangential acceleration also cannot be calculated because we would need their angular accelerations:

\[ a_t=r\alpha \]

Knowing the angular speed at one instant does not tell us how quickly that angular speed is changing.

This question tests an important assumption: equal periods give equal angular speeds only when the motion is uniform.

Question 4

A \(0.50\text{ kg}\) object is attached to a horizontal rotating platform at a radius of \(0.40\text{ m}\).

The platform rotates uniformly with a period of \(2.0\text{ s}\).

A stationary laboratory observer and an observer rotating with the platform describe the object’s motion.

a. Calculate the tangential speed of the object as measured by the laboratory observer.

b. Calculate the magnitude of its inward radial net force.

c. Describe the object’s velocity and acceleration relative to the rotating observer.

d. Explain how the rotating observer accounts for the inward real force without claiming that the object accelerates relative to the platform.

Solution 4

The laboratory observer measures a tangential speed of \(1.26\text{ m s}^{-1}\) and an inward radial net force of \(1.97\text{ N}\). The rotating observer measures zero velocity and zero acceleration relative to the platform, and uses an outward centrifugal pseudo-force to account for the rotating reference frame.

Part a: Tangential speed

First calculate angular speed:

\[ \omega=\frac{2\pi}{2.0} =\pi\ \mathrm{rad\,s^{-1}} \]

Then:

\[ v=\omega r \]

\[ v=\pi(0.40) \]

\[ \boxed{v=1.26\ \mathrm{m\,s^{-1}}} \]

Part b: Inward radial net force

Calculate radial acceleration:

\[ a_r=\omega^2r \]

\[ a_r=\pi^2(0.40) \]

\[ a_r=3.95\ \mathrm{m\,s^{-2}} \]

Using Newton’s second law:

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

\[ F_{\mathrm{net},r} =(0.50)(3.95) \]

\[ \boxed{F_{\mathrm{net},r}=1.97\text{ N}} \]

The force acts towards the centre.

Part c: Rotating observer’s measurements

The object is fixed relative to the platform.

Therefore, its velocity relative to the rotating observer is zero:

\[ v'=0 \]

Its position relative to that observer also remains constant, so:

\[ a'=0 \]

Part d: Reconciling the two descriptions

The rotating observer is using a non-inertial reference frame.

To apply Newton’s laws in this frame, an outward centrifugal pseudo-force is introduced.

Its magnitude is:

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

\[ F_{\mathrm{cf}} =(0.50)(\pi^2)(0.40) \]

\[ \boxed{F_{\mathrm{cf}}=1.97\text{ N}} \]

This balances the real inward force from the platform’s attachment.

There is no contradiction between the two descriptions.

The laboratory observer sees inward acceleration because of a real inward resultant force. The rotating observer sees no relative acceleration because the real inward force is balanced by the centrifugal pseudo-force in that non-inertial frame.

The trap is assuming that an object must have the same velocity and acceleration in every reference frame. It does not.

Question 5

A horizontal turntable rotates uniformly in a clockwise direction with a period of \(2.0\text{ s}\).

Two blocks are attached to the turntable using radial arms.

Their properties are:

Both blocks are fixed relative to the turntable and complete each revolution together.

At a particular instant, both radial arms are released simultaneously. The turntable continues rotating, but assume that horizontal friction and air resistance are negligible. The vertical normal force continues to balance each block’s weight.

Answer the following:

a. Before release, calculate the tangential speed and radial acceleration of each block.

b. Calculate the radial net force on each block. Explain why the forces are equal despite their different speeds and accelerations.

c. A student sitting on the turntable claims that both blocks have zero acceleration and therefore require no inward net force. Explain why this statement is incomplete.

d. Immediately after release, describe the motion of each block as seen by a stationary laboratory observer.

e. Calculate the kinetic energy of each block immediately after release. Explain why equal radial net forces before release do not mean equal kinetic energies afterwards.

Solution 5

Before release, both blocks experience the same inward radial net force of approximately \(1.58\text{ N}\), even though B moves twice as fast and has twice the radial acceleration. After release, each block travels along its own tangent at its existing speed. Block B has twice the kinetic energy of A.

Part a: Tangential speeds and radial accelerations

First calculate the common angular speed:

\[ \omega=\frac{2\pi}{2.0} =\pi\ \mathrm{rad\,s^{-1}} \]

For block A:

\[ v_A=\omega r_A \]

\[ v_A=\pi(0.20) \]

\[ \boxed{v_A=0.628\ \mathrm{m\,s^{-1}}} \]

For block B:

\[ v_B=\pi(0.40) \]

\[ \boxed{v_B=1.26\ \mathrm{m\,s^{-1}}} \]

B moves twice as fast as A.

Now calculate their radial accelerations.

For A:

\[ a_{r,A}=\omega^2r_A \]

\[ a_{r,A}=\pi^2(0.20) \]

\[ \boxed{a_{r,A}=1.97\ \mathrm{m\,s^{-2}}} \]

For B:

\[ a_{r,B}=\pi^2(0.40) \]

\[ \boxed{a_{r,B}=3.95\ \mathrm{m\,s^{-2}}} \]

B also has twice the radial acceleration.

Part b: Radial net forces

For block A:

\[ F_A=m_Aa_{r,A} \]

\[ F_A=(0.80)(1.97) \]

\[ \boxed{F_A=1.58\text{ N}} \]

For block B:

\[ F_B=m_Ba_{r,B} \]

\[ F_B=(0.40)(3.95) \]

\[ \boxed{F_B=1.58\text{ N}} \]

The radial net forces are equal.

We can understand why without repeating the full calculations:

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

Both blocks have the same angular speed.

For A:

\[ m_Ar_A=(0.80)(0.20)=0.16 \]

For B:

\[ m_Br_B=(0.40)(0.40)=0.16 \]

Their mass-radius products are equal, so their radial net force magnitudes are equal.

The tempting prediction is that B must have a greater force because it moves faster.

However, B has only half the mass. Its greater acceleration is exactly balanced by its smaller mass when calculating the net force.

Part c: The rotating observer’s claim

The student is correct that both blocks have zero acceleration relative to the turntable.

However, this does not mean that no inward real force is required.

The student is observing the motion from a rotating, non-inertial reference frame.

Before release, the inward real force from each radial arm is balanced by an outward centrifugal pseudo-force.

Both pseudo-forces have magnitude:

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

For each block, this is approximately \(1.58\text{ N}\).

The laboratory observer describes the blocks as accelerating towards the centre. The rotating observer describes them as stationary relative to the platform.

Both descriptions are consistent when the reference frame is identified correctly.

Part d: Motion immediately after release

Immediately after release, both blocks continue moving along their instantaneous tangential directions.

Their speeds do not suddenly change:

\[ v_A=0.628\ \mathrm{m\,s^{-1}} \]

\[ v_B=1.26\ \mathrm{m\,s^{-1}} \]

With no horizontal resultant force, each block continues in a straight line at constant velocity.

Neither block flies directly outwards from the centre.

The inward force previously supplied by the radial arm has disappeared. Without that force, there is no longer a centripetal acceleration to maintain circular motion.

A person on the rotating platform may describe the blocks as moving away from their original positions relative to the platform. That is a frame-dependent description, not evidence of a new outward real force.

Part e: Kinetic energies

Kinetic energy is calculated using:

\[ E_k=\frac{1}{2}mv^2 \]

For A:

\[ E_{k,A} =\frac{1}{2}(0.80)(0.628)^2 \]

\[ \boxed{E_{k,A}=0.158\text{ J}} \]

For B:

\[ E_{k,B} =\frac{1}{2}(0.40)(1.26)^2 \]

\[ \boxed{E_{k,B}=0.316\text{ J}} \]

B has twice the kinetic energy of A.

This is because B has half the mass but twice the speed. Since kinetic energy depends on the square of speed, doubling speed has a greater effect than halving mass.

Finally, notice that the original period of \(2.0\text{ s}\) no longer describes either block’s motion after release. They are no longer travelling in circular paths.

The central lesson is that period, radius, mass, speed, acceleration, and force are connected, but they are not interchangeable. You must identify the actual constraints before deciding which quantities are equal.

Once you can calculate the required radial net force, the next step is to identify which real interaction supplies it. A free-body diagram will help you decide whether that force comes from tension, friction, gravity, a normal force, or a magnetic interaction.