PhysicsCore24 min read

Circular and Rotational Motion

Angular quantities, centripetal force and the moment of inertia

This topic appears in:

1st Year Physics · ch.3A Level Physics (9702) · ch.12
01

Angular quantities

Definition

Radian — The angle subtended at the centre of a circle by an arc equal in length to the radius. 2π radians make a full turn.

Describing something going round in terms of how far it has travelled along the arc is awkward, because the answer depends on the radius. Every point on a spinning disc sweeps the same angle in the same time, though, so angle is the natural measure.

Angle is measured in radians. One radian is the angle subtended at the centre by an arc equal in length to the radius, which makes the definition θ = s/r. A full turn is 2π radians, so 360° = 2π rad and one radian is about 57.3°.

The radian is not an arbitrary unit chosen to be awkward. Because it is defined as a ratio of two lengths it has no dimensions, and that is what allows s = rθ, v = rω and a = rα to be written without any conversion factor. In degrees every one of those would need a clumsy π/180.

Angular velocity ω is the rate at which the angle changes, in radians per second. Angular acceleration α is the rate at which ω changes.

θ = s / rω = θ / t = 2π / Tv = r ωa = r αθ in radians throughout — these relations are only this simple in radians
θ
anglerad
ω
angular velocityrad s⁻¹
T
periods
v
linear speedm s⁻¹
r
radiusm
02

Why circular motion is accelerated motion

An object going round a circle at a perfectly steady speed is accelerating the entire time, and this is the idea the whole chapter turns on.

Velocity is a vector. Going round a circle changes the direction of motion continuously, so the velocity changes continuously even though its magnitude does not. A changing velocity is, by definition, an acceleration.

The direction of that acceleration is towards the centre. It is called centripetal acceleration, from the Latin for "centre seeking". At every instant the velocity is along the tangent and the acceleration is at right angles to it, pointing inwards.

Because there is an acceleration there must be a resultant force, also directed at the centre. The centripetal force is not a new kind of force. It is whatever real force happens to be doing the job: gravity for a planet, tension for a stone on a string, friction for a car on a bend, the normal contact force for a wall of death rider.

a = v² / r = ω² rF = m v² / r = m ω² rboth a and F point at the centre of the circle
a
centripetal accelerationm s⁻²
F
centripetal forceN
v
linear speedm s⁻¹
r
radiusm

The amber arrow is the velocity, always along the tangent. The blue arrow is the acceleration, always straight at the centre. They stay at right angles no matter where the object is — which is precisely why the speed can stay constant while the velocity changes.

There is no outward force

Nothing pushes you outwards in a turning car. Your body carries on in a straight line while the car turns beneath you, and the door pushes you inwards onto the new path. Writing "centrifugal force" in an answer loses the mark; the only force is centripetal.

03

Working with centripetal force

Every centripetal problem is answered the same way. Identify what is physically providing the inward force, write that force equal to mv²/r, and solve.

For a car on a flat bend the force is friction, so the maximum safe speed comes from setting friction equal to mv²/r. Notice that the mass cancels: a loaded lorry and an empty car skid at the same speed on the same bend, which surprises most people.

For a stone whirled on a string in a vertical circle, the tension and the weight both act along the radius, but they point the same way only at the top. At the top, T + mg = mv²/r; at the bottom, T − mg = mv²/r. The string is therefore slackest at the top and tightest at the bottom, and the minimum speed to keep the string taut at the top is found by setting T = 0.

For a satellite the force is gravity, which is why the orbital speed depends on the radius and not on the mass of the satellite.

Worked example 16 marks

A car of mass 1200 kg takes a flat bend of radius 45 m. The maximum frictional force between the tyres and the road is 8400 N. Find the maximum speed, and state what happens to it if the car is loaded with passengers.

  1. Friction supplies the centripetal force, so F = mv²/r.Identify the real force first — that is the marked step.
  2. 8400 = 1200 v² / 45.
  3. v² = 8400 × 45 / 1200 = 315.
  4. v = 17.7 m s⁻¹, about 64 km/h.
  5. The limiting friction is itself proportional to the weight, so it rises with mass in the same proportion.
  6. The mass cancels and the maximum speed is unchanged.A heavier car does not skid at a lower speed on the same surface.

17.7 m s⁻¹, and loading the car does not change it

04

Rotational motion and moment of inertia

Definition

Moment of inertia — The rotational equivalent of mass: a measure of how hard it is to change a body's rate of rotation. I = Σmr².

Every quantity in linear motion has a rotational twin, and the equations look identical once you swap them over. Force becomes torque, mass becomes moment of inertia, and acceleration becomes angular acceleration.

Moment of inertia depends not only on how much mass there is but on where that mass sits. Mass far from the axis contributes far more, because the contribution goes as r². A hoop and a disc of the same mass and radius have very different moments of inertia, and the hoop is much harder to spin up.

This is why a figure skater spins faster on pulling their arms in. No torque acts, so angular momentum L = Iω is conserved; pulling the arms in reduces I, so ω must rise to keep the product constant. The same physics governs a diver tucking to rotate faster and a neutron star spinning hundreds of times a second after collapsing.

LinearRotationalRelation
displacement sangle θs = rθ
velocity vangular velocity ωv = rω
acceleration aangular acceleration αa = rα
mass mmoment of inertia II = Σmr²
force F = matorque τ = Iα
momentum p = mvangular momentum L = Iω
KE = ½mv²KE = ½Iω²
I = Σ m r²τ = I αL = I ωKE = ½ I ω²angular momentum is conserved when no external torque acts
I
moment of inertiakg m²
τ
torqueN m
L
angular momentumkg m² s⁻¹

Key points

  1. Radians make the angular equations simple — always convert before using them.
  2. Circular motion at constant speed is still accelerated motion.
  3. Centripetal force points at the centre and is supplied by a real force.
  4. Identify that real force first; then set it equal to mv²/r.
  5. Angular momentum Iω is conserved when no torque acts.

Practice questions

6 questions · 25 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Explain why an object moving in a circle at constant speed is accelerating.
Model answer

Its direction of motion is changing continuously, so its velocity is changing. Velocity is a vector, and a changing velocity is an acceleration.

Examiner tip. The word "vector" is the mark. Without it the answer looks like a contradiction.

SQ2[2 marks]
A stone on a string is whirled in a horizontal circle. State what provides the centripetal force, and what happens if the string breaks.
Model answer

The tension in the string. If it breaks there is no longer a resultant force, so the stone flies off along the tangent in a straight line, not outwards along the radius.

Examiner tip. Tangentially, not radially. Drawing the stone flying outwards is the classic error.

SQ3[2 marks]
Explain why a figure skater spins faster when she pulls her arms in.
Model answer

No external torque acts, so angular momentum L = Iω is conserved. Pulling her arms in moves mass closer to the axis, reducing the moment of inertia, so the angular velocity must increase.

Examiner tip. Name the conserved quantity. "Because she is smaller" scores nothing.

Solved numericals

1 · 5 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[5 marks]
A satellite orbits the Earth in a circle of radius 7.0 × 10⁶ m with a period of 5800 s. Calculate its angular velocity, its linear speed, and its centripetal acceleration.
Full working
  1. Uses ω = 2π/T[1]
  2. ω = 2π / 5800 = 1.08 × 10⁻³ rad s⁻¹[1]
  3. Uses v = rω[1]
  4. v = 7.0 × 10⁶ × 1.08 × 10⁻³ = 7.6 × 10³ m s⁻¹[1]
  5. a = ω²r = (1.08 × 10⁻³)² × 7.0 × 10⁶ = 8.2 m s⁻²close to g, as expected in low orbit[1]

ω = 1.08 × 10⁻³ rad s⁻¹, v = 7.6 km s⁻¹, a = 8.2 m s⁻²

Examiner tip. Work in radians throughout. The acceleration coming out near 9.8 is a good sign you have not slipped a power of ten.

Long questions

1 · 8 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[8 marks]
A small ball of mass 0.25 kg is attached to a string of length 0.80 m and swung in a vertical circle.
  1. Draw and label the forces acting on the ball at the top of the circle. [2]
  2. Calculate the minimum speed at the top for the string to stay taut. [3]
  3. Calculate the tension at the bottom if the ball is moving at 6.0 m s⁻¹ there. [3]
Mark scheme
  1. Weight mg acting downwards[1]
  2. Tension T also acting downwards, towards the centreat the top both point the same way[1]
  3. At minimum speed the string just goes slack, so T = 0 and gravity alone provides the force[1]
  4. mg = mv²/r, so v² = gr = 9.81 × 0.80the mass cancels[1]
  5. v = 2.8 m s⁻¹[1]
  6. At the bottom the tension acts up and the weight down: T − mg = mv²/r[1]
  7. T = 0.25 × 6.0² / 0.80 + 0.25 × 9.81[1]
  8. T = 11.25 + 2.45 = 13.7 N[1]

(b) 2.8 m s⁻¹ (c) 13.7 N

Examiner tip. At the top the tension and weight both point inwards, so they add. At the bottom they oppose, so they subtract. Getting the sign the wrong way round is the whole difficulty of this question.

Exam questions

1 · 6 marks

Multi-part questions with a full mark scheme.

Q1[6 marks]
A flywheel of moment of inertia 0.45 kg m² is spinning at 120 rad s⁻¹.
  1. Calculate its rotational kinetic energy. [2]
  2. Calculate its angular momentum. [2]
  3. A ring is dropped onto it, raising the total moment of inertia to 0.60 kg m². Find the new angular velocity. [2]
Mark scheme
  1. Uses KE = ½Iω²[1]
  2. = ½ × 0.45 × 120² = 3240 J[1]
  3. Uses L = Iω[1]
  4. = 0.45 × 120 = 54 kg m² s⁻¹[1]
  5. No external torque, so angular momentum is conserved: 54 = 0.60 ω[1]
  6. ω = 90 rad s⁻¹kinetic energy has fallen — the collision is inelastic[1]

3240 J, 54 kg m² s⁻¹, 90 rad s⁻¹

Examiner tip. Angular momentum is conserved here but kinetic energy is not. Check: the new KE is 2430 J, so 810 J went into heat and sound as the ring gripped.