Uniform circular motion | ICSE Class 10 Physics Notes

This note covers uniform circular motion, speed and velocity, inward acceleration, centripetal force, tangential motion on release, centrifugal force, distance and displacement, and the distinction between inward force and outward reaction.

What is uniform circular motion?

Definition: Uniform circular motion is motion along a circular path at constant speed. Speed tells us how fast an object moves. The word uniform describes the speed throughout this motion.

A circular path is a path around a fixed centre at a fixed distance from it. That distance is the radius. A complete journey round the circle is one revolution, and the length of its boundary is its circumference.

Both conditions in the definition matter. The path must be circular, and the speed must remain constant. Merely saying that an object goes round a circle does not establish that its circular motion is uniform.

What does the word uniform tell us?

Uniform speed means that equal lengths of the path are covered in equal time intervals, however short those intervals are. A time interval is the duration between two instants. An instant refers to a particular moment.

The object keeps changing its direction of motion as it travels round the circle. Thus, uniform circular motion combines something that stays the same, its speed, with something that changes, its direction. These two observations must be kept together.

A child on a moving merry-go-round provides a way to picture the circular path. To treat the child's motion as uniform circular motion, the child's speed along that path must also remain constant.

How closely do real motions fit the model?

An idealised model is a simplified description that treats selected conditions as exact. In the real world, the conditions for uniform circular motion, constant speed and a circular path, are often not met.

The model remains useful for understanding more complex motions, such as planets revolving around the Sun or a vehicle making a circular turn. Calling it a useful model does not mean that every such motion satisfies both conditions exactly.

When identifying the motion, first ask about the shape of the path and then about the speed. A statement about one does not answer the other. The distinction is essential when describing what stays the same and what changes during the motion.

How do speed and velocity differ on a circle?

Velocity describes both how fast an object moves and its direction of motion. Its magnitude is its size, expressed by a numerical value with a unit. At an instant, the magnitude of velocity is the speed.

A scalar quantity needs a magnitude but no direction for its description. A vector quantity needs both magnitude and direction. Speed is a scalar; velocity is a vector. Equal speeds therefore do not establish equal velocities.

Which part of velocity changes?

In uniform circular motion, the magnitude of velocity stays constant. Its direction changes continuously as the object follows the circular path. Since direction is part of velocity, the velocity changes even though the speed remains the same.

The SI unit of speed is the metre per second, written m/s. Here SI means the International System of Units, m denotes metre, and s denotes second. Velocity uses the same unit, with a direction included in its description.

How does a circular track reveal the change?

Imagine the direction followed by an athlete at successive positions on a track. A rectangular track requires turns at its corners. A hexagonal track, which has six sides, requires more frequent turns. A circular track requires a continuous change in direction.

A tangent is a straight line touching a circle at one point. At any point on the circular path, velocity lies along that tangent in the direction of motion.

The figure shows a rectangular track, a hexagonal track and a circular track. Blue arrows labelled v, meaning velocity, indicate directions along the straight sides or along tangents to the circle.

See Fig. 4.23 in your NCERT textbook

The circular drawing is especially useful because its arrows point in different directions around the same path. Read their directions as well as their lengths. Describing the speed alone leaves out the very feature that distinguishes circular motion from straight motion at constant velocity.

Why is uniform circular motion accelerated?

Acceleration is the rate of change of velocity with time. A rate of change describes how much a quantity changes in a given time interval. A change in velocity can result from a change in magnitude, direction, or both.

The SI unit of acceleration is the metre per second squared, written m/s². This unit expresses a change of velocity, measured in metres per second, per second of elapsed time. Acceleration, like velocity, also has a direction.

Why does constant speed not imply zero acceleration?

1. Uniform circular motion has constant speed, so the magnitude of velocity remains the same.
2. The object continuously changes its direction as it travels around the circular path.
3. A change in direction is a change in velocity because velocity is a vector quantity.
4. The continuously changing velocity means that acceleration is present throughout the motion.

There is therefore no contradiction between constant speed and acceleration. The apparent contradiction comes from using the everyday meaning of acceleration, speeding up, in place of its physical meaning, changing velocity.

Note: We often fail to recognise acceleration when only the direction of velocity changes. Checking speed alone is insufficient: the direction of motion must also be considered.

Where does the acceleration point?

The acceleration in uniform circular motion points towards the centre. It is called centripetal acceleration, meaning centre-seeking acceleration. Its direction is inward along a radius, while the velocity is along the direction of travel.

For a fixed circular path at constant speed, the magnitude of this acceleration remains constant. Its direction, however, changes as the object moves. The line from the object to the centre points in a different direction at different positions.

Thus, the acceleration is not a constant vector. Saying that it is always towards the centre does not mean that it points in one fixed direction throughout a revolution. The centre stays fixed, but the moving object's position relative to that centre changes.

This distinction explains why uniform circular motion is accelerated motion without being motion at constant velocity. It also prevents a second mistake: confusing constant acceleration magnitude with constant acceleration, which would require an unchanged direction as well.

How should velocity and acceleration be drawn?

A tangent is a straight line that touches a circle at one point. The velocity of an object moving around the circle lies along the tangent at its current position, in the direction of motion.

The radius drawn to the point of contact is perpendicular to the tangent. Perpendicular means meeting at a right angle. Consequently, the tangential velocity and inward centripetal acceleration are perpendicular to each other in uniform circular motion.

How do the two arrows differ?

A velocity arrow answers, “In which direction is the object moving now?” An acceleration arrow answers, “In which direction is its velocity changing now?” These are different questions, so the arrows need not point in the same direction.

A circle touches the horizontal straight line AC at its upper point B. A and C label positions on that line, B labels the contact point, and a dot marks the centre.

See Fig. 4.25 in your NCERT textbook

The line AC represents the tangent, not a radius. To decide which way a velocity arrow should point along it, the direction of motion around the circle must also be specified. A tangent line alone has no arrow indicating that choice.

How can you construct a complete motion diagram?

Draw a circle with centre O and mark a point P on its boundary. O and P are position labels. Draw the radius OP, a velocity arrow tangent to the circle at P, and an acceleration arrow from P towards O. Label the arrows in words.

1. Mark the circular path and identify its centre before drawing the arrows.
2. Choose and show the direction of motion around the circle.
3. Draw the velocity arrow along the tangent in that direction of motion.
4. Draw the acceleration arrow towards the centre and check that it is perpendicular to the velocity.

A useful check is to imagine the object at another position on the same circle. Its speed is unchanged, but both arrow directions must be reconsidered there. Copying the same fixed arrow directions round the path would misrepresent the motion.

Why is a centripetal force necessary?

A force is a push or pull arising from an interaction. A net force is the combined effect of the forces acting on an object, taking their directions into account. A change in velocity requires a net force.

Centripetal force is the inward force that provides the acceleration needed for circular motion. In uniform circular motion, the net force points towards the centre, in the same direction as the centripetal acceleration.

What does Newton's first law imply?

Newton's first law states that an object remains at rest or continues moving with constant velocity in a straight line unless a net external force changes that state. An external force is a force exerted by something outside the object being considered.

Inertia is the tendency to resist a change in the state of rest or uniform straight-line motion. It does not mean a tendency to continue turning. A circular path requires a continuing change in the direction of velocity.

At constant speed, an object going round a circle still needs an inward force. Without that force, its velocity would not keep turning towards the next part of the circle. The force accounts for the changing direction, rather than an increase in speed.

Is centripetal force a separate kind of force?

Centripetal describes the role and direction of the force. It is not an additional type of interaction to be added to the forces already present. Identify which actual interaction supplies the inward force in the situation.

The SI unit of force is the newton, written N. Centripetal force is measured in this same unit because it is a force. Its special name does not introduce a different unit or a separate physical interaction.

Note: If a string supplies the inward pull, that pull is the centripetal force. Counting the string's pull and a separate centripetal force would count the same inward effect twice.

To analyse a circular motion, name the moving object, identify the inward interaction acting on it, and state the direction of that interaction. This connects the force to the acceleration and avoids treating “centripetal” as an unexplained extra force.

Which interactions can provide centripetal force?

Different interactions can play the same inward role. Tension is the pulling force transmitted by a stretched string. Friction is a contact force opposing relative sliding or its tendency between surfaces. Gravitational attraction is the attractive interaction between objects due to their mass, the measure of inertia.

The question is not simply whether tension, friction or gravity exists. It is whether the interaction provides the force directed towards the centre of the particular circular motion being considered.

How do the familiar examples compare?

For the stone, the string transmits the pull towards the centre. For the car on a horizontal road, friction provides the inward force for turning. For the planet, gravitational attraction provides the inward interaction.

Static friction is friction acting without relative sliding at the contact. In the car example, static friction between the tyres and road supplies the centripetal force while the tyres maintain their grip.

Why must the conditions be stated?

The road being horizontal is part of the car example. It identifies the situation in which friction supplies the inward force. Keeping such conditions attached to an example is more precise than memorising an unrestricted statement about every possible turn.

Similarly, a circular model of planetary motion is an idealisation. The example identifies the interaction responsible for turning the motion; it does not establish that every planet follows an exact circle at exactly constant speed.

These examples show why centripetal force is a useful shared name. The physical interaction can change from one situation to another, but the requirement remains an inward force associated with the circular path. Naming the interaction explains how that requirement is met.

What happens when the inward constraint is removed?

A constraint is something that restricts the motion an object can follow. A ring can constrain a moving marble to follow its inner boundary. Removing the ring removes the contact that was keeping the marble on that curved path.

At the instant of release, the marble already has a velocity along the tangent. It continues in that instantaneous direction when the inward constraint is removed, rather than continuing to follow the circle.

What does the marble activity show?

1. Place a ring, such as an adhesive tape ring, flat on a smooth surface.
2. Set a marble moving along the inner boundary of the ring.
3. After one or two complete revolutions, lift the ring without disturbing the marble's motion.
4. Observe the marble moving in a straight line after release and repeat the activity to confirm the result.

The ring changes the direction of the moving marble while they are in contact. Once the ring is lifted, it can no longer keep directing the marble round its inner boundary. The outgoing direction is the direction the marble had at release.

Why is the release direction tangential?

Tangential motion means motion along the tangent. The marble's velocity just before release is tangential, so there is no need for a new outward push to account for its initial outgoing direction.

Distinguish a tangent from an outward radius. A tangent is perpendicular to the radius at the release point. An outward radius points directly away from the centre. These are different directions, even though a tangentially released object moves away from the circular path.

For continued straight-line motion at constant velocity, the net force after release must be zero. On a real surface, friction can affect the later motion. The activity establishes the release direction without requiring friction to be completely absent.

Note: Inertia preserves the instantaneous straight-line motion when no net force changes it. It does not preserve a circular path. Continuing round the circle would require the velocity direction to keep changing.

When explaining the observation, connect three ideas: the velocity before release, the removal of the inward contact force, and the outgoing tangent. Saying only that the marble “moves away” leaves its direction unclear.

What is centrifugal force, and how does it differ from centripetal force?

A frame of reference is the viewpoint, together with its reference positions, used to describe motion. A rotating frame turns with the circular motion. The distinction between inward and apparent outward force must be tied to the viewpoint being used.

Centrifugal force is the apparent outward force introduced when describing motion from a rotating frame. It points away from the centre. It is also called a pseudo force, a force introduced because the reference frame accelerates, rather than because another object exerts it.

A fixed, non-accelerating frame is called an inertial frame. In its description of uniform circular motion, the object accelerates inward. Its real forces therefore have an inward resultant, meaning an inward net force.

How do the two descriptions compare?

Why can a rider feel an outward effect?

On a merry-go-round, a rider can feel as though they are being pushed outward even though the force maintaining the circular motion acts inward. The feeling of an outward effect must be distinguished from the direction of the force producing the turn.

For an object at rest relative to a uniformly rotating frame, the centrifugal pseudo force balances the inward real force in that rotating description. Relative rest means no change of position as seen from that frame.

This does not mean that the real inward force has disappeared. From the inertial viewpoint, the object is moving round a circle and has centripetal acceleration. Mixing the two viewpoints would incorrectly make its inward net force vanish.

Whenever using the word centrifugal, state that it is the apparent outward force in the rotating description. Keep the force that actually bends the path directed inward. That distinction prevents an outward sensation from being mistaken for an outward cause of circular motion.

How do outward reaction, distance and a complete revolution fit together?

Is centrifugal force the reaction to centripetal force?

Newton's third law states that interacting bodies exert equal and opposite forces on each other. These forces act simultaneously on different bodies. They are often called an action-reaction pair; the names do not imply that one occurs before the other.

For a stone attached to a string, the string pulls the stone inward, and the stone pulls the string outward. The outward pull on the string is a real reaction. It acts on the string, not as an opposing real force on the stone.

The centrifugal pseudo force used in a rotating frame is not that third-law partner. Identifying which body receives each force prevents the real outward reaction on the string from being used to cancel the inward force on the stone.

Does returning to the start mean that no distance was covered?

Distance travelled is the total length of the path covered. Displacement is the net change in position, directed from the initial position to the final position. After one complete revolution, the distance is the circumference, while displacement is zero.

The SI unit of distance is the metre. The SI unit of time is the second. Let R denote the radius and T the time period, meaning the time for one complete revolution. Let π denote the ratio of a circle's circumference to its diameter.

The diameter is the straight-line distance across the circle through its centre. The circumference is 2πR. Average speed is total distance divided by elapsed time, whereas average velocity is displacement divided by elapsed time.

Using v for the constant speed, $v = \frac{2\pi R}{T}$. Over a full revolution the average velocity is zero, because the displacement is zero. The speed, however, is not zero: the object has travelled all the way round the circle.

Zero displacement over a complete revolution does not mean the object was at rest during it. Likewise, zero average velocity over that interval does not remove its instantaneous tangential velocity or the centripetal acceleration present while it moves.

How can circular motion be calculated?

Use metres for radius, seconds for time and kilograms for mass. Let $R$ be radius, $T$ the time period, $v$ the constant speed and $a_c$ the magnitude of the inward acceleration. Convert the given quantities before substituting.

Derivation: How is speed related to the time period?

1. In one complete revolution, the object covers the circumference: $s = 2\pi R$, where $s$ is distance travelled.
2. Speed is distance divided by time. For a revolution completed in time $T$, $v = \frac{s}{T}$.
3. Substitute the circumference for the distance to obtain $v = \frac{2\pi R}{T}$. Since speed is constant, this is also the instantaneous speed.

Result: $v = \frac{2\pi R}{T}$. The period is the time for one revolution, not the total time for several revolutions.

Derivation: Why is centripetal acceleration speed squared divided by radius?

Consider two nearby positions separated by a short time $\Delta t$. Let $|\Delta\mathbf{r}|$ be the chord displacement and $|\Delta\mathbf{v}|$ the magnitude of the velocity change. The speed $v$ and radius $R$ remain constant.

1. The two radius vectors have equal lengths, and the two velocity vectors have equal magnitudes. Each velocity is perpendicular to its radius, so the angle between the velocities equals the angle between the radii.
2. The triangles formed by the radius vectors and by the velocity vectors are therefore similar: $\frac{|\Delta\mathbf{v}|}{v} = \frac{|\Delta\mathbf{r}|}{R}$.
3. Rearrange and divide by the elapsed time: $\frac{|\Delta\mathbf{v}|}{\Delta t} = \frac{v}{R}\frac{|\Delta\mathbf{r}|}{\Delta t}$.
4. As the interval becomes vanishingly small, the chord approaches the short arc, and chord displacement divided by time approaches the speed. The acceleration magnitude becomes $a_c = \frac{v}{R}v = \frac{v^2}{R}$. The velocity change points towards the centre in this limit.

Result: $a_c = \frac{v^2}{R}$, directed towards the centre. Constant speed gives a constant acceleration magnitude, but the acceleration direction keeps changing.

How are frequency, angular speed and force related?

Frequency, written $\nu$, is the number of revolutions per second: $\nu = \frac{1}{T}$. It is measured in hertz. Angular speed, written $\omega$, measures the angle swept out per second, in radians per second.

A full revolution is $2\pi$ radians, giving $\omega = 2\pi\nu$. An arc of length $\Delta s$ subtends an angle $\Delta\theta$ in radians, with $\Delta s = R\Delta\theta$. Dividing by time gives $v = R\omega$.

Substitution in the acceleration formula gives $a_c = R\omega^2$, or $a_c = 4\pi^2\nu^2R$. For mass $m$, Newton's second law gives the inward force magnitude $F_c = ma_c = \frac{mv^2}{R}$.

How do we solve numerical questions?

Worked example 1. An insect moves steadily in a circular groove of radius 12 cm, completing 7 revolutions in 100 s. Find its angular speed, linear speed and acceleration magnitude. Is its acceleration vector constant?

Formula: $\nu = \frac{n}{t}$, $\omega = 2\pi\nu$, $v = R\omega$, $a_c = R\omega^2$. Here $n$ is the number of revolutions and $t$ is the total time.

Substitute: $R = 0.12\,\mathrm{m}$, $\nu = \frac{7}{100} = 0.07\,\mathrm{Hz}$, and $\omega = 2\pi\times0.07 \approx 0.440\,\mathrm{rad/s}$.

Then $v = 0.12\times0.439823 \approx 0.0528\,\mathrm{m/s}$ and $a_c = 0.12\times(0.439823)^2 \approx 0.0232\,\mathrm{m/s^2}$.

Answer: Angular speed is about 0.44 rad/s, speed is 5.3 cm/s, and acceleration magnitude is 2.3 cm/s². The acceleration vector is not constant because its inward direction changes continuously.

Worked example 2. A stone on a string 80 cm long moves at constant speed in a horizontal circle, making 14 revolutions in 25 s. Find the magnitude and direction of its acceleration.

Formula: $v = \frac{2\pi Rn}{t}$, $a_c = \frac{v^2}{R}$, where $n$ revolutions take total time $t$.

Substitute: $R = 0.80\,\mathrm{m}$, so $v = \frac{2\pi\times0.80\times14}{25} \approx 2.815\,\mathrm{m/s}$. Hence $a_c = \frac{(2.814867)^2}{0.80} \approx 9.90\,\mathrm{m/s^2}$.

Answer: The acceleration is about 9.9 m/s² towards the centre of the circle.

Worked example 3. An aircraft flies a horizontal loop of radius 1.00 km at a steady 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Formula: $a_c = \frac{v^2}{R}$. Use $g = 9.8\,\mathrm{m/s^2}$ for the gravitational acceleration.

Substitute: $R = 1000\,\mathrm{m}$, $v = 900\times\frac{1000}{3600} = 250\,\mathrm{m/s}$, and $a_c = \frac{250^2}{1000} = 62.5\,\mathrm{m/s^2}$.

Compare using $\frac{a_c}{g} = \frac{62.5}{9.8} \approx 6.38$.

Answer: The centripetal acceleration is 62.5 m/s² towards the loop's centre, about 6.4 times the acceleration due to gravity.

Worked example 4. Three girls skate along different paths from a point on the edge of a circular ice ground to the diametrically opposite point. The radius is 200 m. What is each girl's displacement magnitude, and when does it equal the path length?

Formula: $d = 2R$, where $d$ is the straight-line separation of the opposite endpoints.

Substitute: $d = 2\times200 = 400\,\mathrm{m}$.

Answer: Each girl's displacement magnitude is 400 m. It equals the distance travelled for the girl skating directly along the straight diameter without reversing direction; any indirect path is longer.

Worked example 5. A stone of mass 0.25 kg is whirled in a horizontal circle of radius 1.5 m at 40 revolutions per minute. Find the string tension and the maximum speed if the string withstands 200 N.

Formula: $v = 2\pi R\nu$, $F = \frac{mv^2}{R}$. Here $F$ denotes the tension supplying the inward force. Rearranging gives $v_{\max} = \sqrt{\frac{F_{\max}R}{m}}$.

Substitute: $\nu = \frac{40}{60}\,\mathrm{Hz}$, so $v = 2\pi\times1.5\times\frac{40}{60} \approx 6.283\,\mathrm{m/s}$.

Then $F = \frac{0.25\times(2\pi)^2}{1.5} \approx 6.58\,\mathrm{N}$, and $v_{\max} = \sqrt{\frac{200\times1.5}{0.25}} = \sqrt{1200}\,\mathrm{m/s}$.

Answer: The tension is about 6.6 N and the maximum speed is about 34.6 m/s.

Worked example 6. A cyclist travels at 18 km/h on a level road and takes a circular turn of radius 3 m without slowing. The coefficient of static friction between tyres and road is 0.1. Will the cyclist slip?

The coefficient of static friction, $\mu_s$, is the ratio of maximum static friction to normal reaction. Friction must supply the required inward force. On a level road the no-slip condition is $v^2 \leq \mu_s Rg$.

Substitute: $v = 18\times\frac{1000}{3600} = 5\,\mathrm{m/s}$, giving $v^2 = 25\,\mathrm{m^2/s^2}$. But $\mu_s Rg = 0.1\times3\times9.8 = 2.94\,\mathrm{m^2/s^2}$.

Answer: At 5 m/s, the cyclist slips because 25 exceeds 2.94 in the squared-speed comparison. The available static friction cannot provide the required centripetal force.

Glossary

• Uniform circular motion — Motion along a circular path with speed remaining constant throughout the motion.
• Speed — The magnitude of velocity, describing how fast an object moves without specifying its direction.
• Velocity — A quantity describing both the speed of an object and its direction of motion.
• Acceleration — The rate of change of velocity with time, including changes in direction.
• Tangent — A straight line touching a circle at a single point on its boundary.
• Centripetal acceleration — The acceleration directed towards the centre during uniform circular motion.
• Centripetal force — The inward force that supplies the acceleration required for circular motion.
• Centrifugal force — An apparent outward force introduced when motion is described from a rotating reference frame.
• Inertia — The tendency of an object to resist changes in rest or uniform straight-line motion.
• Tension — The pulling force transmitted through a stretched string to an attached object.
• Displacement — The net change in position, directed from an object's initial position to its final position.
• Time period — The time taken by an object to complete one revolution around its circular path.

Common errors and misconceptions

• Misconception: Constant speed means constant velocity. Correct: Velocity also includes direction. Its direction changes continuously during uniform circular motion, although its magnitude remains constant.
• Misconception: Acceleration requires speeding up. Correct: A change in velocity direction also produces acceleration. Uniform circular motion has inward acceleration while its speed remains unchanged.
• Misconception: Velocity points towards the centre. Correct: Velocity is tangential; centripetal acceleration and the net force point inward. Their different directions describe different aspects of motion.
• Misconception: Centripetal force is added to the string's tension. Correct: When tension supplies the inward force, it performs the centripetal role. Do not count that force twice.
• Misconception: A released marble moves directly along an outward radius. Correct: Its initial release direction is tangential, following the velocity it had when the ring was removed.
• Misconception: An outward reaction cancels the inward force on the stone. Correct: The stone pulls the string outward, while the string pulls the stone inward. These forces act on different bodies.
• Misconception: Returning to the starting point means zero distance travelled. Correct: Displacement is zero after a complete revolution, but distance travelled equals the circumference of the circular path.

Exam-style questions with model answers

Q1. Define uniform circular motion and state what remains constant in it. [2 marks]

1. Uniform circular motion is the motion of an object along a circular path at constant speed.
2. The speed, which is the magnitude of velocity, remains constant throughout the motion.

Q2. An object moves at constant speed around a circle. Explain why it is accelerated and state the direction of its acceleration. [3 marks]

1. Velocity depends on both speed and direction. Following a circular path continuously changes the object's direction of motion, even though its speed stays constant.
2. Acceleration is the rate of change of velocity. The changing direction therefore gives the object acceleration without requiring it to speed up.
3. The acceleration is directed towards the centre of the circle and is called centripetal acceleration.

Q3. A marble moves around the inner boundary of a ring on a smooth horizontal surface. The ring is lifted without disturbing the marble. State its initial release direction and explain why it does not continue round the circle. [3 marks]

1. The marble initially moves along the tangent to the circle at the release point. This is its direction of velocity immediately before the ring is lifted.
2. Lifting the ring removes the inward contact force that was continually changing the direction of the marble's velocity.
3. Without that inward constraint, the marble does not keep following the circular boundary. Its initial outgoing direction is tangential, rather than directly outward along a radius.

Q4. Compare centripetal force with centrifugal force in a rotating frame. Give three differences, then explain the inward force and apparent outward effect for a rider moving uniformly with a merry-go-round. [5 marks]

1. Centripetal force points towards the centre of the circular path, whereas centrifugal force in a rotating frame points away from that centre.
2. Centripetal force is supplied by an actual interaction. Centrifugal force is a pseudo force introduced because the viewpoint rotates and accelerates.
3. In an inertial description, centripetal force explains the inward acceleration. Centrifugal force is not added as another real force on the moving rider.
4. The force maintaining the rider's circular motion acts inward, continuously changing the direction of velocity even though the speed stays constant.
5. The rider can nevertheless feel an outward effect. In the rotating description, the centrifugal pseudo force accounts for that apparent effect without replacing the real inward interaction.

Q5. A stone moves uniformly around a circle while attached to a stretched string. Name the centripetal interaction, identify its third-law partner, and explain why the two forces do not cancel on the stone. [3 marks]

1. Tension in the string provides the inward force on the stone. This pull supplies the centripetal force needed to change the stone's velocity direction.
2. The third-law partner is the equal and opposite outward force exerted by the stone on the string.
3. The inward force acts on the stone, while its partner acts on the string. They therefore do not cancel as forces on the stone itself.

Q6. A child makes one complete revolution on a merry-go-round at constant speed. The circular path has radius R and the revolution takes time T. Here R is a length, T is a time, and π is the ratio of circumference to diameter. Find the distance travelled, displacement, average speed and average velocity. [4 marks]

1. The distance travelled is the circumference, 2πR, because the child completes the whole circular path once.
2. The displacement is zero because the final position coincides with the starting position after the revolution.
3. The average speed is total distance divided by time, giving 2πR/T. This also equals the constant speed during the motion.
4. The average velocity is displacement divided by time. Since displacement is zero over this interval, the average velocity is zero.

Key takeaways

• Uniform circular motion requires a circular path and constant speed; the object's direction changes continuously as it moves.
• Velocity includes direction as well as magnitude, so constant speed around a circle does not mean constant velocity.
• Centripetal acceleration points towards the centre, while the instantaneous velocity points along the tangent to the circular path.
• The inward force may be supplied by tension, friction or gravity; centripetal describes its role rather than a separate interaction.
• A marble released from a circular constraint initially follows the tangent, continuing in the direction of its velocity at release.
• Centrifugal force is an apparent outward force used in a rotating frame, distinct from the real inward force causing circular motion.
• The stone's outward pull on a string is a real reaction acting on the string, not an extra outward force on the stone.
• One complete revolution gives zero displacement but a non-zero distance equal to the circumference, so average velocity and average speed differ.

Test yourself

Which two conditions define uniform circular motion?

The object must move along a circular path and maintain constant speed throughout that motion.

What changes when speed stays constant around a circle?

The direction of velocity changes continuously. Since direction is part of velocity, the velocity changes and acceleration is present.

Where should you draw the velocity arrow at a point on the circle?

Draw it along the tangent at that point, pointing in the direction in which the object is moving.

Does inward acceleration point in one fixed direction throughout a revolution?

No. It points from the object's changing position towards the centre, so its direction changes continuously even though its magnitude stays constant.

What supplies the centripetal force for a car turning on a horizontal road without slipping?

Static friction between the tyres and the road supplies the inward force needed for the circular turn.

Why does a marble released from a ring initially follow a tangent?

Its velocity is tangential at release. Removing the ring removes the inward constraint, so its initial outgoing motion follows that existing direction.

On which object does the reaction to the string's inward pull on a stone act?

It acts on the string: the stone pulls the string outward. It therefore does not cancel the inward force on the stone.

How can average velocity be zero over a revolution when speed is not zero?

The object returns to its starting position, giving zero displacement. It has still covered the circumference, so its average speed is non-zero.