Motion in a circle

Both subtopics here are printed under "A Level subject content" in the 9702 syllabus, and every objective carries the tier "A Level". None of it is AS. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper, and the whole 9702 question bank on this site comes from Paper 1, so there is no practice tagged to this topic.

Syllabus points

12.1 Kinematics of uniform circular motion

• Define the radian and express angular displacement in radians.
• Understand and use the concept of angular speed.
• Recall and use ω = 2π/T and v = rω.

12.2 Centripetal acceleration

• Understand that a force of constant magnitude that is always perpendicular to the direction of motion causes centripetal acceleration.
• Understand that centripetal acceleration causes circular motion with a constant angular speed.
• Recall and use a = rω² and a = v²/r.
• Recall and use F = mrω² and F = mv²/r.

The radian

One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius.

θ = s / r

where s is the arc length and r the radius. Because a full circle has circumference 2πr, a complete revolution is 2π radians, so 2π rad = 360°.

To convert:

180 / π = 57.3

so one radian is about 57.3°. Working in radians is not a convenience here, it is a requirement: every formula below assumes it, and a calculator left in degree mode is the single commonest way to lose all the marks on a question you understood.

Worked example. A point on the rim of a wheel of radius 0.40 m moves through an arc of 0.30 m.

θ = 0.30 / 0.40 = 0.75

so the angular displacement is 0.75 rad.

Angular speed

Angular speed ω is the rate of change of angular displacement:

ω = θ / t

measured in rad s⁻¹. For one complete revolution the object turns through 2π radians in a time T, the period, so:

ω = 2π / T

and since frequency f = 1/T, ω = 2πf.

The linear speed of a point moving in a circle is related to ω by:

v = rω

Read that as a proportionality and it explains a great deal. Every point on a rotating rigid body has the same angular speed, because the whole body turns through the same angle in the same time. But the linear speed increases with radius. A point on the rim of a record moves faster than a point near the spindle, even though both complete a revolution in the same time. Questions comparing two points on the same rotating object are almost always testing exactly this.

Worked example. A wheel of radius 0.35 m rotates at 45 revolutions per minute.

Convert to a period first:

T = 60 / 45 = 1.333

so T = 1.33 s.

ω = 2 × 3.142 / 1.333 = 4.714

so ω = 4.71 rad s⁻¹, and the speed of a point on the rim is

v = 0.35 × 4.714 = 1.650

so v = 1.65 m s⁻¹.

Why circular motion needs a force

Velocity is a vector. An object moving round a circle at constant speed is continually changing direction, so its velocity is continually changing, so it is accelerating, even though the speedometer reading never moves.

An acceleration requires a resultant force. For the speed to stay constant, the force must have no component along the direction of motion, which means it must be perpendicular to the velocity at every instant. A force perpendicular to the velocity does no work, since work is force times distance moved in the direction of the force, and there is no displacement in that direction. That is why the kinetic energy, and therefore the speed, is unchanged.

Perpendicular to the velocity, for motion in a circle, means along the radius towards the centre. Hence the names: the acceleration is centripetal, meaning centre-seeking, and so is the force.

If a force of constant magnitude is always perpendicular to the motion, the result is motion in a circle at constant angular speed, which is what uniform circular motion means.

Centripetal acceleration

a = v² / r

and substituting v = rω gives the other form:

a = rω²

Both are on the formula sheet and each suits different data. Note that they pull in opposite directions with respect to r, and this catches people out:

• At constant speed v, a smaller radius means a larger acceleration. Take a bend tighter and you need more grip.
• At constant angular speed ω, a larger radius means a larger acceleration. On a spinning turntable the outer coin flies off first.

There is no contradiction. In the first case v is fixed and ω must change; in the second ω is fixed and v must change. Always check which quantity the question is holding constant.

Centripetal force

From F = ma:

F = mv² / r

F = mrω²

This is the point at which the topic is most often misunderstood, so state it carefully. Centripetal force is not a new kind of force. It is the name given to the resultant force that happens to be acting towards the centre, and in any real situation some identifiable force is providing it:

A question that asks "what provides the centripetal force" wants the physical agent, not the words "centripetal force".

There is no outward force. The sensation of being thrown outwards in a turning car is inertia: your body continues in a straight line while the car turns beneath you, and the door pushes you inwards. Writing "centrifugal force" into an answer loses marks.

Worked example. A car of mass 1200 kg takes a bend of radius 50 m at 15 m s⁻¹.

F = 1200 × 15² / 50 = 5400

so the centripetal force required is 5400 N, and it must be supplied by friction. If the maximum frictional force available is less than this, the car cannot stay on the circle and slides outwards on a wider path.

The maximum speed follows from the maximum friction. If the tyres can supply at most 7500 N:

v² = 7500 × 50 / 1200 = 312.5

v = 17.7 m s⁻¹

Note that mass cancels when the friction is proportional to weight, which is why the safe speed on a bend does not depend much on how heavily the car is loaded.

The vertical circle

A mass on a string swung in a vertical circle is the standard harder problem. The tension and the weight both act along the radius at the top and the bottom, so:

At the top, both the tension and the weight point towards the centre, which is downwards:

T + mg = mv² / r

At the bottom, the tension points towards the centre, upwards, and the weight points away:

T - mg = mv² / r

So the tension is greatest at the bottom and least at the top, which is why a string breaks at the bottom of the swing.

The minimum speed at the top is the speed at which the string goes slack, T = 0, so gravity alone provides the centripetal force:

mg = mv² / r, so v² = gr

Worked example. A bucket of water is swung in a vertical circle of radius 0.80 m. What is the minimum speed at the top for the water to stay in?

v² = 9.81 × 0.80 = 7.848

v = 2.80 m s⁻¹

Below that, the required centripetal force is less than the weight, gravity is more than sufficient, and the water leaves the bucket.

Common mistakes

• Leaving the calculator in degree mode. Every equation in this topic assumes radians.
• Saying an object moving at constant speed in a circle is not accelerating. Its velocity changes because its direction changes.
• Treating centripetal force as an extra force acting alongside the real ones, and adding it into a free-body diagram.
• Writing "centrifugal force" as an outward force on the object.
• Saying the centripetal force does work on the object. It is perpendicular to the motion, so it does none, and the speed is unchanged.
• Using a = v²/r and a = rω² without noticing which of v and ω the question holds constant, then concluding that acceleration both rises and falls with radius.
• Assuming two points on the same rotating body have the same linear speed. They share ω, not v.
• Adding the weight to the tension at the bottom of a vertical circle instead of subtracting it.
• Forgetting to convert revolutions per minute to a period in seconds before finding ω.