Torque and Rotational Dynamics

So far we have studied objects moving in circles. Now we turn to objects that rotate — spinning about an axis. The physics of rotation mirrors the physics of linear motion, but with angular quantities replacing linear ones. This lesson covers torque, moment of inertia, angular momentum, and conservation of angular momentum.

Torque

You already know that a moment is force × perpendicular distance from the pivot. A torque (or the torque of a couple) specifically refers to a pair of equal and opposite forces that cause rotation without translation.

For a single force at perpendicular distance d from the axis of rotation:

$\tau = Fd$

Where τ (Greek letter tau) is the torque, measured in N m.

More generally, if the force is at angle θ to the line joining the point of application to the axis:

$\tau = Fr\sin\theta$

Couples

A couple consists of two equal and opposite forces separated by a distance d. The torque of a couple is:

$\tau = Fd$

where F is the magnitude of one of the forces and d is the perpendicular distance between them. A couple produces pure rotation with no net translational force.

Worked Example 1

A spanner of length 0.25 m is used to tighten a bolt. A force of 60 N is applied perpendicular to the spanner. What torque is applied?

$\tau = Fd = 60 \times 0.25 = 15 \text{ N m}$

If the force is applied at 30° to the spanner (rather than perpendicular):

$\tau = Fr\sin\theta = 60 \times 0.25 \times \sin 30° = 60 \times 0.25 \times 0.50 = 7.5 \text{ N m}$

Linear–Rotational Analogies

Linear Quantity	Symbol	Unit	Rotational Equivalent	Symbol	Unit
Displacement	s	m	Angular displacement	θ	rad
Velocity	v	m s⁻¹	Angular velocity	ω	rad s⁻¹
Acceleration	a	m s⁻²	Angular acceleration	α	rad s⁻²
Mass (inertia)	m	kg	Moment of inertia	I	kg m²
Force	F	N	Torque	τ	N m
Momentum	p = mv	kg m s⁻¹	Angular momentum	L = Iω	kg m² s⁻¹
Newton's 2nd law	F = ma			τ = Iα
KE	½mv²	J	Rotational KE	½Iω²	J
Impulse	FΔt = Δp		Angular impulse	τΔt = ΔL

Moment of Inertia

In linear motion, mass measures how hard it is to change an object's velocity (inertia). In rotational motion, the equivalent quantity is the moment of inertia (I), which measures how hard it is to change an object's angular velocity.

$I = \sum m_i r_i^2$

The moment of inertia depends on both the mass and how that mass is distributed relative to the axis of rotation. Mass further from the axis contributes more to the moment of inertia.

Common Moments of Inertia

Object	Axis	Moment of Inertia	Example
Point mass (m) at distance r	Through centre	mr²	Ball on string
Solid cylinder/disc (M, R)	Through centre	½MR²	Flywheel, CD
Hollow cylinder (M, R)	Through centre	MR²	Bicycle wheel rim
Solid sphere (M, R)	Through centre	⅖MR²	Bowling ball
Hollow sphere (M, R)	Through centre	⅔MR²	Football
Thin rod (M, L)	Through centre	¹⁄₁₂ML²	Ruler balanced at centre
Thin rod (M, L)	Through one end	⅓ML²	Bat swung from handle

Worked Example 2

A flywheel can be modelled as a solid disc of mass 8.0 kg and radius 0.20 m. What is its moment of inertia?

$I = \frac{1}{2}MR^2 = \frac{1}{2} \times 8.0 \times 0.20^2 = \frac{1}{2} \times 8.0 \times 0.04 = 0.16 \text{ kg m}^2$

Worked Example 3: Compound Object

A grinding wheel consists of a solid disc (mass 4.0 kg, radius 0.15 m) mounted on an axle (solid cylinder, mass 1.0 kg, radius 0.02 m). Find the total moment of inertia.

$I_{\text{disc}} = \frac{1}{2} \times 4.0 \times 0.15^2 = 0.045 \text{ kg m}^2$ $I_{\text{axle}} = \frac{1}{2} \times 1.0 \times 0.02^2 = 0.0002 \text{ kg m}^2$ $I_{\text{total}} = 0.045 + 0.0002 = 0.0452 \text{ kg m}^2$

The axle contributes negligibly because its radius is so small — almost all the rotational inertia is in the disc.

Newton's Second Law for Rotation

The rotational equivalent of F = ma is:

$\tau = I\alpha$

Where τ is the net torque, I is the moment of inertia, and α is the angular acceleration (in rad s⁻²).

Worked Example 4

The flywheel (I = 0.16 kg m²) has a torque of 2.4 N m applied to it. What is its angular acceleration?

$\alpha = \frac{\tau}{I} = \frac{2.4}{0.16} = 15 \text{ rad s}^{-2}$

Starting from rest, after 4.0 s its angular velocity would be: $\omega = \alpha t = 15 \times 4.0 = 60 \text{ rad s}^{-1}$

Angular Momentum

Just as linear momentum is p = mv, angular momentum is:

$L = I\omega$

Angular momentum is measured in kg m² s⁻¹ (or equivalently, N m s).

For a point mass moving in a circle: $L = I\omega = mr^2 \times \frac{v}{r} = mvr$

Conservation of Angular Momentum

In the absence of an external torque, the total angular momentum of a system is conserved:

$I_1\omega_1 = I_2\omega_2$

This is the rotational equivalent of conservation of linear momentum.

The Ice Skater

The most famous demonstration of conservation of angular momentum is the spinning ice skater. When the skater pulls their arms in:

Their moment of inertia decreases (mass is closer to the axis)
Their angular velocity increases (to conserve angular momentum)
The spin speeds up dramatically

Worked Example 5

An ice skater spins at 3.0 rev s⁻¹ with arms extended (I = 4.0 kg m²). They pull their arms in, reducing their moment of inertia to 1.5 kg m². What is their new spin rate?

$I_1\omega_1 = I_2\omega_2$ $4.0 \times 3.0 = 1.5 \times \omega_2$ $\omega_2 = \frac{12}{1.5} = 8.0 \text{ rev s}^{-1}$

The skater speeds up from 3 to 8 revolutions per second.

Energy analysis: KE = ½Iω².

Before: KE = ½ × 4.0 × (3.0 × 2π)² = ½ × 4.0 × 355.3 = 711 J

After: KE = ½ × 1.5 × (8.0 × 2π)² = ½ × 1.5 × 2527 = 1895 J

The kinetic energy increases by 1184 J. This energy comes from the work done by the skater's muscles pulling their arms inward against the centripetal acceleration.

Other Examples of Angular Momentum Conservation

flowchart TD
    A["Conservation of\nAngular Momentum\nI₁ω₁ = I₂ω₂"] --> B["Ice skater\nArms in → smaller I → faster spin"]
    A --> C["Diver/gymnast\nTuck → smaller I → faster rotation\nLayout → larger I → slower rotation"]
    A --> D["Neutron star\nCollapse → tiny R → enormous ω\nSome spin 700+ rev s⁻¹"]
    A --> E["Turntable + clay\nDrop mass on → I increases\n→ ω decreases"]
    A --> F["Earth–Moon system\nMoon receding → r increases\n→ Earth’s spin slows"]

Worked Example 6: Turntable and Clay

A turntable (I = 0.020 kg m²) spins at 3.5 rad s⁻¹. A 0.10 kg lump of clay is dropped onto it at radius 0.15 m. What is the new angular velocity?

$I_{\text{clay}} = mr^2 = 0.10 \times 0.15^2 = 0.00225 \text{ kg m}^2$ $I_{\text{total}} = 0.020 + 0.00225 = 0.02225 \text{ kg m}^2$ $0.020 \times 3.5 = 0.02225 \times \omega_2$ $\omega_2 = \frac{0.07}{0.02225} = 3.15 \text{ rad s}^{-1}$

Worked Example 7: Collapsing Star

A star of radius 7.0 × 10⁸ m rotates once every 30 days. It collapses to a neutron star of radius 10 km = 10⁴ m. Find its new rotation period. (Model as uniform sphere: I = ⅖MR²)

$I_1\omega_1 = I_2\omega_2$ $\frac{2}{5}MR_1^2 \times \frac{2\pi}{T_1} = \frac{2}{5}MR_2^2 \times \frac{2\pi}{T_2}$ $R_1^2 / T_1 = R_2^2 / T_2$ $T_2 = T_1 \times \frac{R_2^2}{R_1^2} = 30 \times 86400 \times \frac{(10^4)^2}{(7 \times 10^8)^2}$ $T_2 = 2.592 \times 10^6 \times \frac{10^8}{4.9 \times 10^{17}} = 2.592 \times 10^6 \times 2.04 \times 10^{-10}$ $T_2 = 5.29 \times 10^{-4} \text{ s}$

The neutron star rotates about 1890 times per second — a millisecond pulsar.

Common Exam Mistakes

Mistake 1: Confusing torque (N m) with energy (J). They have the same dimensions but are fundamentally different quantities. Torque is a vector; energy is a scalar.

Mistake 2: Assuming rotational KE is conserved when angular momentum is conserved. When a skater pulls their arms in, L is conserved but KE increases (the skater does work).

Mistake 3: Forgetting that moment of inertia depends on the axis. A rod has I = ¹⁄₁₂ML² about its centre but I = ⅓ML² about one end.

Summary

Torque: τ = Fd (for perpendicular force) or τ = Fr sin θ (general)
Moment of inertia: I = Σmr² — depends on mass distribution relative to axis
Newton's second law for rotation: τ = Iα
Angular momentum: L = Iω
Conservation of angular momentum: Iω = constant (if no external torque)
Ice skater: arms in → smaller I → faster spin (angular momentum conserved)
Rotational KE = ½Iω²; this is NOT conserved when I changes (even though L is)

A-Level Deep Dive: Torque and Rotational Dynamics

Spec mapping

Edexcel 9PH0 specification Topic 6 — Further Mechanics covers angular displacement, angular velocity $\omega$ , the relationship $v = r\omega$ , centripetal acceleration $a = r\omega^2 = v^2/r$ , and the analysis of forces in circular motion (refer to the official specification document for exact wording). The rotational-dynamics extension — torque $\tau$ , moment of inertia $I$ , the rotational form of Newton's second law $\tau = I\alpha$ , angular momentum $L = I\omega$ and its conservation — sits at the synoptic intersection of Topic 6 with Topic 3 (Working as a Physicist), Topic 4 (Mechanics) and the Year 2 Topic 13 (Oscillations). Although the bare formula $F = mv^2/r$ appears in the Edexcel formula booklet, the rotational analogues $\tau = I\alpha$ and $L = I\omega$ are not tabulated and must be reproduced from first principles by analogy with linear mechanics. Torque is also examined in Topic 8 (Electric and Magnetic Fields) when motors and current-carrying loops in magnetic fields are analysed, and underpins the gyroscopic stabilisation arguments that appear in the Engineering option papers.

Worked example with full mark scheme

Question (8 marks):

A uniform solid disc of mass $M = 2.0\,\text{kg}$ and radius $R = 0.30\,\text{m}$ is mounted on a frictionless horizontal axle through its centre. A light inextensible string is wrapped around the rim and a mass $m = 0.50\,\text{kg}$ hangs from the free end. The mass is released from rest. The moment of inertia of the disc about the axle is $I = \tfrac{1}{2}MR^2$ .

(a) Show that the linear acceleration $a$ of the falling mass is given by $a = \dfrac{mg}{m + \tfrac{1}{2}M}$ . (5)

(b) Hence calculate the angular acceleration $\alpha$ of the disc and the tension $T$ in the string. (3)

Solution with mark scheme:

(a) Step 1 — apply Newton's second law to the falling mass.

$mg - T = ma$

M1 — correct free-body equation for the mass, with weight $mg$ acting downward and tension $T$ upward; net force equals $ma$ . A common slip is to write $T - mg = ma$ (wrong sign convention), which loses M1 immediately.

Step 2 — apply the rotational form of Newton's second law to the disc.

The string applies a tangential force $T$ at radius $R$ , producing a torque $\tau = TR$ . Hence

$TR = I\alpha = \tfrac{1}{2}MR^2 \alpha$

M1 — using $\tau = I\alpha$ with the correct moment of inertia for the disc.

Step 3 — link linear and angular kinematics.

Because the string does not slip, the linear acceleration of the rim equals the linear acceleration of the falling mass: $a = R\alpha$ , so $\alpha = a/R$ .

M1 — explicit no-slip constraint $a = R\alpha$ .

Step 4 — substitute.

From Step 2: $TR = \tfrac{1}{2}MR^2 \cdot (a/R) = \tfrac{1}{2}MRa$ , so $T = \tfrac{1}{2}Ma$ .

Substitute into Step 1: $mg - \tfrac{1}{2}Ma = ma$ , giving $mg = (m + \tfrac{1}{2}M)a$ .

A1 — algebra leading to the printed expression.

A1 — final result $a = \dfrac{mg}{m + \tfrac{1}{2}M}$ presented exactly as requested.

(b) Numerical evaluation:

$a = \dfrac{0.50 \times 9.81}{0.50 + 1.0} = \dfrac{4.905}{1.5} = 3.27\,\text{m s}^{-2}$

B1 — $a \approx 3.3\,\text{m s}^{-2}$ .

$\alpha = a/R = 3.27/0.30 = 10.9\,\text{rad s}^{-2}$ . B1 — $\alpha \approx 11\,\text{rad s}^{-2}$ .

$T = \tfrac{1}{2}Ma = 0.5 \times 2.0 \times 3.27 = 3.27\,\text{N}$ . B1 — $T \approx 3.3\,\text{N}$ .

Total: 8 marks (M3 A2 B3, split as shown). Note the consistency check: $T < mg = 4.9\,\text{N}$ , as required for the mass to accelerate downward.

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): An ice skater of moment of inertia $I_1 = 4.5\,\text{kg m}^2$ spins at $\omega_1 = 2.0\,\text{rev s}^{-1}$ with arms outstretched. She pulls her arms in, reducing her moment of inertia to $I_2 = 1.8\,\text{kg m}^2$ .

(a) Calculate her new angular velocity $\omega_2$ , stating the principle used. (3)

(b) Show that her rotational kinetic energy increases, and explain the physical origin of the additional energy. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.2) — stating conservation of angular momentum: $I_1 \omega_1 = I_2 \omega_2$ , justified by the absence of external torque about the vertical spin axis.
M1 (AO2.1) — substitution: $\omega_2 = (4.5/1.8) \times 2.0 = 5.0\,\text{rev s}^{-1}$ .
A1 (AO1.1b) — answer with units; conversion to $\omega_2 \approx 31\,\text{rad s}^{-1}$ acceptable.

(b)

M1 (AO2.1) — initial KE $= \tfrac{1}{2}I_1\omega_1^2$ , final KE $= \tfrac{1}{2}I_2\omega_2^2$ (using $\omega$ in $\text{rad s}^{-1}$ ).
A1 (AO1.1b) — $\Delta\text{KE}$ positive: ratio of final to initial $= I_2\omega_2^2 / (I_1\omega_1^2) = (I_1/I_2)$ since $\omega_2 = (I_1/I_2)\omega_1$ , giving a factor of $2.5$ .
A1 (AO3.2) — explanation: the skater does internal work against the centripetal force needed to pull her arms inward; this muscular work supplies the extra KE.

Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. Edexcel reserves the AO3 mark for the physical interpretation — candidates who only crunch numbers cap at 5/6.

Synoptic links

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