Rotational Dynamics

Everything you have learned about forces, motion and momentum in a straight line has a rotational twin. This workbook builds that second toolkit — torque, angular acceleration, moment of inertia and angular momentum — so you can analyse spinning and turning bodies, from spanners and seesaws to flywheels and ice skaters, with the same confidence you already have for linear motion.

By the end of this workbook you should be able to:

Calculate the torque of a force about an axis using τ = Fr sin θ, and combine several torques acting on a body.
Explain what it means for a body to be in rotational equilibrium, and use this to solve balance problems.
Describe rotational motion using angular displacement, angular velocity and angular acceleration, and apply the equations of motion for uniform angular acceleration.
Explain what determines the moment of inertia of a body, and calculate it for a system of point masses.
Apply Newton's second law for rotation, τ = Iα, to predict angular acceleration.
Calculate angular momentum (L = Iω) and angular impulse (ΔL = τΔt), and apply conservation of angular momentum.
Calculate the kinetic energy of a rotating body.

Name

1. Torque

In earlier work you studied dynamics — the branch of physics that explains motion in terms of its causes. So far this has mostly meant bodies moving in straight lines: a resultant force produces a linear acceleration. In this topic we study rigid bodies (objects that keep their shape) that can turn about a fixed axis, such as a wheel, a door or a spanner. It turns out that almost every idea you already know from linear motion has a rotational "twin". The table below is worth returning to throughout this workbook.

Linear motion	Rotational motion
force, F	torque, τ
inertial mass, m	moment of inertia, I
displacement, s	angular displacement, θ
speed / velocity, v	angular speed / velocity, ω
acceleration, a	angular acceleration, α
momentum, p = mv	angular momentum, L = Iω
kinetic energy, ½mv²	rotational kinetic energy, ½Iω²

To make something rotate, a force has to act somewhere other than exactly through the axis of rotation. The turning effect of a force is called its torque.

Key idea. The torque of a force about an axis depends on three things: the size of the force, F; the distance from the axis to the point where the force acts, r; and the angle, θ, between the force's line of action and the line joining the axis to the point of application.

τ = F r sin θ

Only the component of the force that is perpendicular to r (that is, F sin θ) actually contributes to turning — a push or pull directed straight along r has no turning effect at all, because it passes through the axis. Torque is measured in newton-metres (N m).

There are two equivalent ways to picture where the sin θ comes from — splitting up the force, or splitting up r. Both lead to exactly the same result.

Fig. 1.1(a) Splitting the force F into a component along r (which passes through the axis and has no turning effect) and a component perpendicular to r, of size F sin θ.

Fig. 1.1(b) Alternatively, r sin θ is the component of r that is perpendicular ("at 90°") to the line of action of F.

Both give exactly the same torque: τ = F(sin θ)r = F(r sin θ) = Fr sin θ.

Line of action: the straight line, extending in both directions, along which a force acts. Axis of rotation: the fixed line about which a body turns. Pivot: a fixed point supporting something that turns or balances.

Careful: torque has the SI unit N m — the same unit as energy (the joule). This is a coincidence of units, not a connection of meaning: torque is a turning effect, not a form of energy.

Worked example 1.1

A cyclist pushes down on a bicycle pedal with a force of 320 N. The pedal crank is 0.17 m long, and the line of action of the force makes an angle of 70° with the crank. Calculate the torque produced about the axle.

Answer:
τ = Fr sin θ
τ = 320 × 0.17 × sin 70°

Worth remembering Torque is greatest when θ = 90°. When the force is applied perpendicular to r, sin θ = 1 and τ = Fr — this is the same "force × perpendicular distance" rule you may already know as the moment of a force. When θ = 0° or 180° (the force acts along the line to the axis), sin θ = 0 and the torque is zero, no matter how big the force is.

Check your understanding

1A mechanic applies a force of 45 N perpendicular to a spanner of length 0.20 m to loosen a nut. Calculate the torque produced.

The force is perpendicular to the spanner, so θ = 90° and sin θ = 1.
τ = Fr sin θ = 45 × 0.20 × sin 90° = 45 × 0.20 × 1

2A force of 60 N is applied to a door handle 0.80 m from the hinge, at an angle of 40° to the door. Calculate the torque produced about the hinge.

τ = Fr sin θ = 60 × 0.80 × sin 40°

3Using τ = Fr sin θ, explain why pushing a door at the very edge (far from the hinge) needs less force to produce the same turning effect than pushing near the hinge.

For a given torque τ, F and r are inversely related (τ = Fr sin θ, so F = τ / (r sin θ)). Increasing r (pushing further from the hinge) means a smaller force F is needed to produce the same torque, provided the angle θ stays the same.

2. Rotational equilibrium

You already know that a body is in translational equilibrium when the resultant force acting on it is zero — it stays still, or carries on moving in a straight line at constant speed. There is a rotational version of the same idea.

Key idea. A body is in rotational equilibrium when the resultant torque acting on it is zero. It is then either not rotating at all, or rotating at a constant angular speed.

Torque, like the moment of a force, has a sense — clockwise or anticlockwise. When several torques act on a body, we choose a positive direction and add them up, treating torques in the opposite sense as negative. For rotational equilibrium:

sum of clockwise torques = sum of anticlockwise torques

If the torques do not balance, there is a resultant (unbalanced) torque, and this will cause the body to undergo an angular acceleration — it will speed up or slow down its rotation. We will return to exactly how much angular acceleration a given torque produces in Section 6.

Worth remembering Rotational equilibrium is not the whole story. A body is in total equilibrium only when both the resultant force and the resultant torque are zero. It is possible for a body to have zero resultant force but a non-zero resultant torque (for example, a steering wheel turned by two equal and opposite forces on either side) — the body won't accelerate in a straight line, but it will still spin up.

Worked example 2.1

A uniform plank is pivoted at its centre. A child of weight 250 N sits 1.5 m from the pivot. A second child of weight 300 N sits on the other side of the pivot. How far from the pivot must the second child sit for the plank to balance?

Answer:
Both weights act vertically (perpendicular to the plank), so θ = 90° for each and sin θ = 1.
For equilibrium: clockwise torque = anticlockwise torque
250 × 1.5 = 300 × x

Live simulation: PhET Balancing Act

Add masses to the see-saw at different distances from the pivot. Before you release each mass, use τ = Fr sin θ (with θ = 90°, since weight acts straight down, perpendicular to the horizontal beam) to predict whether the see-saw will balance — then test your prediction.

Interactive simulation — open the online version of this workbook to launch it.

Simulation by PhET Interactive Simulations, University of Colorado Boulder.

Check your understanding

1A uniform beam is pivoted at its centre. A force of 40 N acts vertically downward 0.60 m from the pivot on one side. What vertical force, acting 0.48 m from the pivot on the other side, is needed for equilibrium?

Both forces act perpendicular to the beam (θ = 90°).
clockwise torque = anticlockwise torque
40 × 0.60 = F × 0.48

2A uniform beam pivoted at its centre has three forces acting on it, all perpendicular to the beam: 20 N clockwise at 0.50 m from the pivot, 15 N clockwise at 0.30 m from the pivot, and an unknown anticlockwise force at 0.80 m from the pivot. Calculate the size of the unknown force needed for equilibrium.

Total clockwise torque = (20 × 0.50) + (15 × 0.30) = 10 + 4.5 = 14.5 N m
For equilibrium, anticlockwise torque = clockwise torque:
F × 0.80 = 14.5

3A spanner is being used to loosen a very tight nut, but pushing on it doesn't seem to work. Suggest two ways to increase the torque applied without increasing the force used.

Since τ = Fr sin θ: (1) use a longer spanner, increasing r; (2) push as close as possible to perpendicular to the spanner (θ closer to 90°), which maximises sin θ.

3. Describing rotational motion

Just as linear motion is described using displacement, velocity and acceleration, rotational motion is described using three closely related quantities.

Angular displacement, θ, is the total angle through which a body has rotated from a fixed reference position. It is measured in radians (rad).

Angular velocity, ω, is the rate of change of angular displacement: ω = Δθ / Δt. Its unit is rad s⁻¹.

Angular acceleration, α, is the rate of change of angular velocity: α = Δω / Δt. Its unit is rad s⁻².

Fig. 3.1 As point P moves along its circular path, it sweeps out an angular displacement θ and travels an arc length s = rθ.

Radian: the angle subtended at the centre of a circle by an arc equal in length to the radius. There are 2π radians in a complete circle (360°). Angular velocities are often quoted in revolutions per minute (rpm); 1 rpm = 2π/60 rad s⁻¹.

All points on a rigid rotating body share the same angular velocity, but points further from the axis move faster in a straight-line (linear) sense, because they must cover a greater distance in the same time. The linear speed v of a point a distance r from the axis is related to ω by:

v = ωr

and, in the same way, the linear (tangential) acceleration a of that point is related to the angular acceleration by a = αr.

Worked example 3.1

A fairground ride completes one full rotation every 4.5 s, moving at a constant rate. Calculate (a) its angular velocity, and (b) the linear speed of a seat 6.0 m from the centre.

Answer:
(a) ω = Δθ/Δt = 2π / 4.5
(b) v = ωr = (2π / 4.5) × 6.0

Check your understanding

1A motor spins at 45 rpm. Show how you would convert this to an angular velocity in rad s⁻¹.

1 rpm = 2π/60 rad s⁻¹, so:
ω = 45 × (2π/60)

2A big wheel rotates at a constant angular velocity, turning through 1.5 complete revolutions in 90 s. Calculate its angular velocity in rad s⁻¹.

Total angle turned, Δθ = 1.5 × 2π rad.
ω = Δθ/Δt = (1.5 × 2π) / 90

4. Equations of motion for angular acceleration

When a body has a uniform (constant) angular acceleration, the equations of motion you already use for straight-line motion work equally well for rotation — you just swap each linear quantity for its rotational partner.

Linear motion (constant a)	Rotational motion (constant α)
v = u + at	ω_f = ω_i + αt
s = ((u + v)/2) t	Δθ = ((ω_i + ω_f)/2) t
s = ut + ½at²	Δθ = ω_it + ½αt²
v² = u² + 2as	ω_f² = ω_i² + 2αΔθ

Key idea. ω_i is the angular velocity at the start of the time interval, and ω_f is the angular velocity at the end. These equations only apply when the angular acceleration α is constant.

Fig. 4.1 Angular velocity–time graph for a rotating body over 10 s.

Worked example 4.1

A CD spins up from rest to 480 rpm in 3.0 s, at a uniform angular acceleration. Calculate (a) the final angular velocity in rad s⁻¹, (b) the angular acceleration, and (c) the angular displacement during this time.

Answer:
(a) ω_f = 480 × (2π/60)
(b) α = (ω_f − ω_i) / t = (ω_f − 0) / 3.0
(c) Δθ = ((ω_i + ω_f)/2) t = ((0 + ω_f)/2) × 3.0

Check your understanding

1A fan blade decelerates uniformly from 12 rad s⁻¹ to rest in 8.0 s. Calculate its angular deceleration.

ω_f = ω_i + αt
0 = 12 + α(8.0)
α = −12/8.0

2Look at Fig. 4.1. State the angular acceleration during the first 4 s, and calculate the total angular displacement over the full 10 s shown.

Angular acceleration = gradient = (8 − 0)/(4 − 0) = 2 rad s⁻².
Total angular displacement = area under the graph = area of trapezium
= (½ × (10 + 4) × 8) [using parallel sides of 10 s and 4 s, height 8 rad s⁻¹]

3An object rotating at 15 rad s⁻¹ accelerates uniformly at 4.0 rad s⁻² for 5.0 s. Calculate (a) its angular displacement during this time, and (b) its final angular velocity.

(a) Δθ = ω_it + ½αt² = (15 × 5.0) + (½ × 4.0 × 5.0²)
(b) ω_f = ω_i + αt = 15 + (4.0 × 5.0)

5. Moment of inertia

Newton's first law tells us that a resultant force is needed to change an object's linear velocity — an object's resistance to this change is called its inertia, and it depends on mass. Rotational motion has its own version of inertia.

Key idea. The moment of inertia, I, of a body measures its resistance to a change in rotational motion (angular acceleration). Unlike mass, it depends not just on how much mass a body has, but on how that mass is distributed relative to the axis of rotation — mass further from the axis contributes more to I than the same mass close to the axis.

The simplest case is a single point mass, m, a distance r from the axis:

I = mr²

For a system made up of several point masses, we add up the contribution of each one:

I = Σmr²

(Σ means "the sum of".) The SI unit of moment of inertia is kg m².

Fig. 5.1 Three bodies of equal mass and radius, rotating about a central axis. The hoop has all its mass concentrated at the rim (largest I); the sphere has most of its mass close to the axis (smallest I).

You do not need to memorise moment of inertia formulas for shapes such as discs, rods or spheres — in an exam, these would be given to you when needed. What matters is understanding what moment of inertia depends on, and being able to use I = Σmr² for systems of point masses.

Fig. 5.2 A system of four point masses fixed to the ends of two light rods, rotating about the axis through their common centre.

Worked example 5.1

Four point masses of 250 g each are fixed to the ends of two light rods that cross at the centre, as in Fig. 5.2. Each mass is 0.45 m from the axis of rotation, which passes through the centre, perpendicular to the page. Calculate the moment of inertia of the system.

Answer:
I = Σmr² = 4 × (0.250 × 0.45²)

Live simulation: the great rolling race

Hoop vs disc: same mass and radius, different moment of inertia

Ramp angle: 25°

Press "Release" to start the race.

Fig. 5.3 A hoop (I = mr²) and a solid disc (I = ½mr²) of equal mass and radius, released from rest at the top of a ramp. Shown in slow motion.

Check your understanding

1A solid sphere and a hollow (thin-walled) sphere have the same mass and the same radius. Which has the larger moment of inertia about a diameter? Explain your answer in terms of the distribution of mass — you do not need a formula.

The hollow sphere has the larger moment of inertia. In the hollow sphere all the mass is concentrated at the outer surface, as far from the axis as possible, while in the solid sphere much of the mass is distributed closer to the centre (closer to the axis). Since moment of inertia depends on how far the mass is from the axis (not just how much mass there is), the hollow sphere resists angular acceleration more.

2A dumbbell consists of two point masses of 400 g, fixed to the ends of a light rod. Each mass is 0.30 m from the axis of rotation, which passes through the centre of the rod. Calculate the moment of inertia of the dumbbell.

I = Σmr² = 2 × (0.400 × 0.30²)

3Using the simulation above, explain why the disc reaches the bottom of the ramp before the hoop, even though they have the same mass and the same radius.

The hoop has all its mass concentrated at its rim, furthest from the axis, so it has a larger moment of inertia than the disc (I = mr² compared with I = ½mr²). A larger moment of inertia means the hoop resists angular acceleration more, for the same torque due to gravity, so it accelerates — and rolls down the ramp — more slowly than the disc.

6. Newton's second law for rotation

For linear motion, Newton's second law tells us that a resultant force F acting on a mass m produces an acceleration a, given by F = ma. Torque, moment of inertia and angular acceleration are related in exactly the same way.

Key idea. Newton's second law for rotation: a resultant (average) torque τ acting on a body of moment of inertia I produces an angular acceleration α.

τ = Iα

This equation tells us that, for a given torque, a body with a larger moment of inertia will have a smaller angular acceleration — it is "harder to spin up" — exactly as a larger mass is harder to accelerate linearly for a given force.

Worked example 6.1

A resultant torque of 18 N m acts on a flywheel with a moment of inertia of 3.6 kg m². Calculate the angular acceleration produced.

Answer:
α = τ / I = 18 / 3.6

Check your understanding

1A torque of 12 N m is applied to a stationary wheel of moment of inertia 2.4 kg m². Calculate (a) the angular acceleration produced, and (b) the angular velocity of the wheel after 5.0 s.

(a) α = τ/I = 12/2.4
(b) ω_f = ω_i + αt = 0 + (α × 5.0), using the value of α found in (a)

2The wheels of a braking car are decelerated by a resultant torque of 320 N m from the brakes. Each wheel has a moment of inertia of 1.8 kg m². Calculate the angular deceleration of a wheel.

α = τ/I = 320/1.8

7. Angular momentum

Linear momentum, p = mv, is one of the most powerful ideas in physics because it is conserved. Rotational motion has an equivalent quantity.

Key idea. The angular momentum, L, of a rotating body is the product of its moment of inertia and its angular velocity.

L = Iω

Angular momentum has the SI unit kg m² s⁻¹.

Conservation of angular momentum. The total angular momentum of a system remains constant, provided that no resultant (external) torque acts on it.

This is why a spinning ice skater speeds up when she pulls her arms in: bringing her mass closer to her axis of rotation reduces her moment of inertia, I. Since no external torque acts on her (ignoring friction), L = Iω must stay constant — so if I decreases, ω must increase to compensate.

Live simulation: conservation of angular momentum

Pull the masses in — watch the spin speed change

Distance of each mass from the axis, r: 0.90 m

r = 0.90 m · I = 2.43 kg m² · ω = 2.00 rad s⁻¹

Fig. 7.1 Two 1.5 kg masses rotate about a central axis. L = Iω is fixed at its starting value — dragging the slider changes r (and so I), and ω updates automatically to keep L constant.

Worth remembering Conservation of angular momentum shows up all over nature. Divers and gymnasts curl into a tight ball to spin faster, then straighten out to slow down before landing. Even neutron stars — the very dense, collapsed remnants of much larger stars — spin extremely fast, because the star's moment of inertia becomes tiny as it collapses, while its angular momentum is conserved.

Worked example 7.1

A solid disc with moment of inertia 0.045 kg m² spins at 6.0 rad s⁻¹. A ring of modelling clay is then dropped onto the disc, increasing the total moment of inertia to 0.060 kg m². Calculate the new angular velocity of the disc-and-clay system.

Answer:
Angular momentum is conserved (no external torque): I₁ω₁ = I₂ω₂
0.045 × 6.0 = 0.060 × ω₂

Check your understanding

1Calculate the angular momentum of an ice skater with a moment of inertia of 1.2 kg m², rotating at 3.5 rad s⁻¹.

L = Iω = 1.2 × 3.5

2An ice skater spinning with her arms outstretched has a moment of inertia of 4.2 kg m² and an angular velocity of 2.0 rad s⁻¹. She pulls her arms in, reducing her moment of inertia to 1.4 kg m². Calculate her new angular velocity.

L is conserved: I₁ω₁ = I₂ω₂
4.2 × 2.0 = 1.4 × ω₂

3Using conservation of angular momentum, explain why a diver curls into a tight ball during a somersault, and then straightens out just before entering the water.

Curling into a ball moves the diver's mass closer to their axis of rotation, reducing their moment of inertia. Since angular momentum L = Iω is conserved (no external torque acts once they leave the board), a smaller I means a larger ω, so they spin faster and can complete more rotations. Straightening out before entry increases I again, reducing ω so they enter the water with a slow, controlled rotation.

8. Angular impulse

You have already met linear impulse, J = FΔt, the change in linear momentum produced by a force acting for a time Δt. There is a rotational equivalent.

Key idea. A resultant torque τ, acting for a time Δt, produces a change in angular momentum called the angular impulse.

ΔL = τΔt = Δ(Iω)

The SI unit of angular impulse is kg m² s⁻¹ (equivalently, N m s).

If the torque is not constant, we use its average value. Just as the area under a force–time graph gives the linear impulse, the area under a torque–time graph gives the angular impulse (the change in angular momentum) — this is true whatever the shape of the graph.

Fig. 8.1 A torque that rises instantly to 5.0 N m, stays constant until t = 2.0 s, then falls steadily to zero by t = 3.0 s.

Worked example 8.1

Determine the angular impulse represented by the torque–time graph in Fig. 8.1.

Answer:
Split the shaded area into a rectangle (0 to 2.0 s) and a triangle (2.0 to 3.0 s).
Angular impulse = area of rectangle + area of triangle
ΔL = (5.0 × 2.0) + (½ × 5.0 × 1.0)

Check your understanding

1A constant torque of 4.5 N m acts on a wheel for 3.0 s. Calculate the angular impulse produced.

ΔL = τΔt = 4.5 × 3.0

2An angular impulse of 12 kg m² s⁻¹ is applied to a wheel of moment of inertia 2.0 kg m², initially at rest. Calculate the final angular velocity of the wheel.

ΔL = Δ(Iω) = Iω_f − Iω_i. Since ω_i = 0:
12 = 2.0 × ω_f

9. Rotational kinetic energy

A moving mass has linear kinetic energy, ½mv². A rotating body has kinetic energy too, because every particle of the body is moving, even though the body as a whole may not be travelling anywhere.

Key idea. The kinetic energy of a rotating body follows directly from the linear-rotational analogy.

E_k = ½Iω² = ½ L²/I

Worth remembering Rolling objects have both kinds of kinetic energy at once. A ball or wheel rolling without slipping is moving forward (translational KE = ½mv²) and spinning (rotational KE = ½Iω²) at the same time, so its total kinetic energy is the sum of both. This is why, in exam questions about a ball rolling down a slope, the loss in gravitational potential energy is shared between linear and rotational kinetic energy — a rolling object always ends up moving slower than one that just slides down frictionlessly.

Worked example 9.1

A flywheel with a moment of inertia of 0.85 kg m² spins at 420 rpm. Calculate its rotational kinetic energy.

Answer:
ω = 420 × (2π/60)
E_k = ½Iω² = ½ × 0.85 × ω² (using the value of ω found above)

Check your understanding

1A rotating disc has a moment of inertia of 0.020 kg m² and an angular velocity of 15 rad s⁻¹. Calculate its rotational kinetic energy.

E_k = ½Iω² = ½ × 0.020 × 15²

2Two flywheels have the same rotational kinetic energy. Flywheel A has twice the moment of inertia of flywheel B. Show that the angular velocity of flywheel A is 1/√2 times the angular velocity of flywheel B.

Equal kinetic energies: ½I_Aω_A² = ½I_Bω_B²
Since I_A = 2I_B: 2I_Bω_A² = I_Bω_B²
Cancel I_B and rearrange: ω_A² = ω_B²/2, so ω_A = ω_B/√2 = (1/√2)ω_B

3A flywheel stores 500 J of rotational kinetic energy while spinning at 25 rad s⁻¹. Calculate its moment of inertia.

E_k = ½Iω²
500 = ½ × I × 25²

Glossary

Torque, τ: The turning effect of a force about an axis: τ = Fr sin θ. Measured in N m.
Line of action: The straight line, extended in both directions, along which a force acts.
Axis of rotation: The fixed line about which a rigid body turns.
Rotational equilibrium: The state of a body when the resultant torque acting on it is zero, so it does not undergo angular acceleration.
Angular displacement, θ: The total angle through which a rigid body has rotated from a reference position, measured in radians.
Angular velocity, ω: The rate of change of angular displacement: ω = Δθ/Δt.
Angular acceleration, α: The rate of change of angular velocity: α = Δω/Δt.
Moment of inertia, I: A measure of a body's resistance to angular acceleration, which depends on its mass and how that mass is distributed relative to the axis of rotation. For a point mass, I = mr²; for a system of point masses, I = Σmr².
Newton's second law for rotation: The resultant (average) torque acting on a body equals its moment of inertia multiplied by its angular acceleration: τ = Iα.
Angular momentum, L: The rotational equivalent of linear momentum: L = Iω. Measured in kg m² s⁻¹.
Conservation of angular momentum: The principle that the total angular momentum of a system stays constant unless a resultant external torque acts on it.
Angular impulse: The change in angular momentum produced by a torque acting for a time interval: ΔL = τΔt = Δ(Iω).
Rotational kinetic energy: The kinetic energy a body has because of its rotation: E_k = ½Iω² = ½ L²/I.