Rotational Dynamics

Having built the moment of inertia, we now write down the dynamics of rotation: how torques produce angular acceleration, how rotating bodies carry kinetic energy, and how energy and work transfer in rolling, braking and pulley systems. The governing law $\tau = I\alpha$τ=Iα is the exact rotational mirror of $F = ma$F=ma, and almost every linear result you know — SUVAT, kinetic energy, work, power — has a rotational twin obtained by the dictionary $s \to \theta$s→θ, $v \to \omega$v→ω, $a \to \alpha$a→α, $m \to I$m→I, $F \to \tau$F→τ. This lesson develops that dictionary carefully and then applies it to the three classic Paper-3 scenarios: a rolling body on a slope, a braked flywheel, and an Atwood machine with a massive pulley.

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1 Where this sits in AQA 7367

This is Paper 3, the Mechanics option (7367/3M), sitting directly on top of the moments-of-inertia lesson. It is a rich source of multi-stage problems that fuse all three assessment objectives: write down $\tau = I\alpha$τ=Iα or an energy equation (AO1), choose between a force/torque approach and an energy approach and justify the rolling condition (AO2), and carry a modelling problem — slope, pulley, brake — to a numerical answer (AO3). Paper 3 weights these AO1 40% / AO2 25% / AO3 35%. The prerequisites are the SUVAT and energy methods of A-Level Mechanics, plus the moment-of-inertia results from the previous lesson.

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2 Core theory — Newton's second law for rotation

The rotational analogue of $F = ma$F=ma is

$$\boxed{\,\tau = I\alpha\,}$$τ=Iα​

where $\tau$τ is the resultant torque about the axis ($\text{N m}$N m), $I$I the moment of inertia about that axis ($\text{kg m}^2$kg m2), and $\alpha = \ddot{\theta}$α=θ¨ the angular acceleration ($\text{rad s}^{-2}$rad s−2). It follows from $\tau = \dfrac{\mathrm{d}L}{\mathrm{d}t}$τ=dtdL​ (next lesson) when $I$I is constant: $\tau = \dfrac{\mathrm{d}(I\omega)}{\mathrm{d}t} = I\dfrac{\mathrm{d}\omega}{\mathrm{d}t} = I\alpha$τ=dtd(Iω)​=Idtdω​=Iα.

Torque

The torque (moment) of a force $F$F about an axis is

$$\tau = Fd = Fr\sin\theta,$$τ=Fd=Frsinθ,

where $d = r\sin\theta$d=rsinθ is the perpendicular distance from the axis to the line of action of the force, $r$r the distance to the point of application, and $\theta$θ the angle between $\mathbf{F}$F and the position vector $\mathbf{r}$r. Only the component of force perpendicular to $\mathbf{r}$r produces a turning effect; a force pointing straight at (or away from) the axis exerts no torque. There are two equivalent ways to read $Fr\sin\theta$Frsinθ: as "force times perpendicular distance to the line of action" ($F\times r\sin\theta$F×rsinθ), or as "perpendicular component of force times distance" ($F\sin\theta\times r$Fsinθ×r). Both give the same number, and being fluent with both readings saves time in problems where one geometry is more obvious than the other. In vector language $\boldsymbol{\tau} = \mathbf{r}\times\mathbf{F}$τ=r×F, and the cross product automatically extracts the $\sin\theta$sinθ and fixes the sense of rotation by the right-hand rule.

The rotational SUVAT equations

When the angular acceleration $\alpha$α is constant, the angular kinematics are identical in form to linear SUVAT:

Rotational	Linear analogue
$\omega = \omega_0 + \alpha t$ω=ω0​+αt	$v = u + at$v=u+at
$\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$θ=ω0​t+21​αt2	$s = ut + \tfrac12 at^2$s=ut+21​at2
$\theta = \omega t - \tfrac{1}{2}\alpha t^2$θ=ωt−21​αt2	$s = vt - \tfrac12 at^2$s=vt−21​at2
$\omega^2 = \omega_0^2 + 2\alpha\theta$ω2=ω02​+2αθ	$v^2 = u^2 + 2as$v2=u2+2as
$\theta = \tfrac{1}{2}(\omega_0 + \omega)t$θ=21​(ω0​+ω)t	$s = \tfrac12(u+v)t$s=21​(u+v)t

with the substitution dictionary $s \to \theta$s→θ, $u \to \omega_0$u→ω0​, $v \to \omega$v→ω, $a \to \alpha$a→α. Crucially these hold only for constant $\alpha$α — exactly as linear SUVAT needs constant $a$a.

Rotational kinetic energy, work and power

$$\text{KE}_{\text{rot}} = \tfrac12 I\omega^2, \qquad W = \tau\theta\ (\text{constant }\tau),\quad W = \int\tau\,\mathrm{d}\theta\ (\text{variable}), \qquad P = \tau\omega .$$KErot​=21​Iω2,W=τθ (constant τ),W=∫τdθ (variable),P=τω.

For a body that both translates and rotates (a rolling wheel, a sphere down a slope) the total kinetic energy splits cleanly into a translational part for the centre of mass and a rotational part about the centre of mass:

$$\boxed{\,\text{KE}_{\text{total}} = \tfrac12 m v_{\text{cm}}^2 + \tfrac12 I_{\text{cm}}\omega^2\,}.$$KEtotal​=21​mvcm2​+21​Icm​ω2​.

For rolling without slipping the contact point is instantaneously at rest, which forces the rolling condition

$$v = R\omega, \qquad a = R\alpha .$$v=Rω,a=Rα.

It is worth understanding why the contact point is at rest. The velocity of any point of a rolling wheel is the sum of the centre's velocity $v$v (forwards) and the rotational velocity $r\omega$rω about the centre. At the very bottom these are equal and opposite ($v$v forwards from translation, $R\omega$Rω backwards from spin), so they cancel when $v = R\omega$v=Rω — the no-slip condition. The top of the wheel, by contrast, moves at $v + R\omega = 2v$v+Rω=2v, twice the centre's speed. This is why a rolling wheel photographs sharply at the bottom and blurred at the top.

A second consequence is energetic: because the contact point is instantaneously stationary, the friction acting there does no work, so mechanical energy is conserved in pure rolling even though friction is essential to cause the rolling. This is the licence to use $mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2$mgh=21​mv2+21​Iω2 without a friction-loss term — a point examiners reward when stated explicitly.

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3 Worked examples with mark scheme

Example 1 — torque on a disc

A torque of $6\,\text{N m}$6N m acts on a disc of moment of inertia $1.5\,\text{kg m}^2$1.5kg m2. Find the angular acceleration.

$$\alpha = \frac{\tau}{I} = \frac{6}{1.5} = 4\ \text{rad s}^{-2}. \quad (\textbf{M1 A1})$$α=Iτ​=1.56​=4 rad s−2.(M1 A1)

Example 2 — constant angular acceleration

A wheel starts from rest and accelerates at $3\,\text{rad s}^{-2}$3rad s−2 for $4\,\text{s}$4s. Find (a) the final angular speed, (b) the angle turned, (c) the number of revolutions.

$$\text{(a)}\ \omega = \omega_0 + \alpha t = 0 + 3(4) = 12\ \text{rad s}^{-1}. \quad (\textbf{M1 A1})$$(a) ω=ω0​+αt=0+3(4)=12 rad s−1.(M1 A1) $$\text{(b)}\ \theta = \omega_0 t + \tfrac12\alpha t^2 = 0 + \tfrac12(3)(4^2) = 24\ \text{rad}. \quad (\textbf{M1 A1})$$(b) θ=ω0​t+21​αt2=0+21​(3)(42)=24 rad.(M1 A1) $$\text{(c)}\ \frac{\theta}{2\pi} = \frac{24}{2\pi} \approx 3.82\ \text{revolutions}. \quad (\textbf{A1})$$(c) 2πθ​=2π24​≈3.82 revolutions.(A1)

(M1 correct SUVAT equation each part; A1 each value. To convert to revolutions divide by $2\pi$2π, not $360$360 — the angles are in radians.)

Example 3 — rolling down a slope

A uniform solid sphere of mass $2\,\text{kg}$2kg, radius $0.1\,\text{m}$0.1m, rolls without slipping from rest down a slope, descending a vertical height $3\,\text{m}$3m. Find its speed at the bottom $(g = 9.8)$(g=9.8).

For a solid sphere $I = \tfrac25 MR^2$I=52​MR2; rolling gives $\omega = v/R$ω=v/R. Conserve energy: (M1)

$$\begin{aligned} mgh &= \tfrac12 mv^2 + \tfrac12 I\omega^2 = \tfrac12 mv^2 + \tfrac12\cdot\tfrac25 mR^2\cdot\frac{v^2}{R^2} \quad (\textbf{M1})\\ mgh &= \tfrac12 mv^2 + \tfrac15 mv^2 = \tfrac{7}{10}mv^2 \\ v^2 &= \frac{10gh}{7} = \frac{10(9.8)(3)}{7} = 42 \ \Rightarrow\ v = \sqrt{42} \approx 6.48\ \text{m s}^{-1}. \quad (\textbf{A1}) \end{aligned}$$mghmghv2​=21​mv2+21​Iω2=21​mv2+21​⋅52​mR2⋅R2v2​(M1)=21​mv2+51​mv2=107​mv2=710gh​=710(9.8)(3)​=42 ⇒ v=42​≈6.48 m s−1.(A1)​

(M1 energy conservation with both KE terms; M1 substitute $I$I and $\omega = v/R$ω=v/R; A1 $v \approx 6.48$v≈6.48. The mass and radius cancel — the answer depends only on $g$g, $h$h and the shape factor $\tfrac25$52​.)

Key insight. A rolling sphere is slower than a frictionless sliding particle ($v = \sqrt{2gh} \approx 7.67\,\text{m s}^{-1}$v=2gh​≈7.67m s−1) because some potential energy becomes rotational KE. A hoop ($I = MR^2$I=MR2) ends up slowest of all ($v = \sqrt{gh}$v=gh​); the larger the shape factor, the slower the descent.

Example 4 — braking a flywheel

A flywheel with $I = 2\,\text{kg m}^2$I=2kg m2 spins at $100\,\text{rad s}^{-1}$100rad s−1. A constant braking torque of $5\,\text{N m}$5N m is applied. How many revolutions does it turn before stopping?

$$\alpha = -\frac{\tau}{I} = -\frac{5}{2} = -2.5\ \text{rad s}^{-2}. \quad (\textbf{M1})$$α=−Iτ​=−25​=−2.5 rad s−2.(M1)

Using $\omega^2 = \omega_0^2 + 2\alpha\theta$ω2=ω02​+2αθ with $\omega = 0$ω=0:

$$0 = 100^2 + 2(-2.5)\theta \ \Rightarrow\ 5\theta = 10000 \ \Rightarrow\ \theta = 2000\ \text{rad}. \quad (\textbf{M1 A1})$$0=1002+2(−2.5)θ ⇒ 5θ=10000 ⇒ θ=2000 rad.(M1 A1) $$\frac{2000}{2\pi} \approx 318\ \text{revolutions}. \quad (\textbf{A1})$$2π2000​≈318 revolutions.(A1)

(M1 $\alpha = -\tau/I$α=−τ/I with the negative sign for deceleration; M1 the right SUVAT; A1 $\theta = 2000$θ=2000; A1 revolutions. An energy route is equally valid: $\tau\theta = \tfrac12 I\omega_0^2 \Rightarrow 5\theta = \tfrac12(2)(100^2) = 10000$τθ=21​Iω02​⇒5θ=21​(2)(1002)=10000, the same $\theta$θ.)

Example 5 — rolling cylinder by forces, and the friction required

A uniform cylinder of mass $m$m and radius $R$R rolls without slipping down a rough plane inclined at $\beta$β to the horizontal. By writing the translational and rotational equations of motion, find the acceleration of its centre and the frictional force, and hence the least coefficient of friction for rolling.

Let $f$f be the friction up the slope and $a = R\alpha$a=Rα. Resolving along the slope and taking moments about the centre: (M1)

$$\begin{aligned} \text{along slope:}\quad & mg\sin\beta - f = ma, \quad (1)\\ \text{about centre:}\quad & fR = I\alpha = \tfrac12 mR^2\cdot\frac{a}{R} = \tfrac12 mRa \ \Rightarrow\ f = \tfrac12 ma. \quad (2)\ (\textbf{M1}) \end{aligned}$$along slope:about centre:​mgsinβ−f=ma,(1)fR=Iα=21​mR2⋅Ra​=21​mRa ⇒ f=21​ma.(2) (M1)​

Substitute (2) into (1): $mg\sin\beta - \tfrac12 ma = ma \Rightarrow mg\sin\beta = \tfrac32 ma$mgsinβ−21​ma=ma⇒mgsinβ=23​ma, so

$$a = \tfrac23 g\sin\beta, \qquad f = \tfrac12 ma = \tfrac13 mg\sin\beta. \quad (\textbf{A1})$$a=32​gsinβ,f=21​ma=31​mgsinβ.(A1)

For rolling (static friction) we need $f \le \mu R_N = \mu mg\cos\beta$f≤μRN​=μmgcosβ:

$$\tfrac13 mg\sin\beta \le \mu mg\cos\beta \ \Rightarrow\ \mu \ge \tfrac13\tan\beta. \quad (\textbf{M1 A1})$$31​mgsinβ≤μmgcosβ ⇒ μ≥31​tanβ.(M1 A1)

(M1 both equations; M1 eliminate $f$f via $a = R\alpha$a=Rα; A1 $a = \tfrac23 g\sin\beta$a=32​gsinβ; M1 friction inequality; A1 $\mu \ge \tfrac13\tan\beta$μ≥31​tanβ. Notice the energy method gives $a$a faster, but only the force method yields $f$f and the friction condition — so the question dictates the approach.)

This worked example shows the forces route that complements the energy route of Example 3: both give $a = \tfrac23 g\sin\beta$a=32​gsinβ for a cylinder, a satisfying cross-check, but only the forces route exposes the friction.

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4 Specimen-style exam question — Atwood machine with a massive pulley

(specimen-style — not from any past paper)

Masses of $5\,\text{kg}$5kg and $3\,\text{kg}$3kg hang from a light inextensible string passing over a pulley modelled as a uniform disc of mass $2\,\text{kg}$2kg and radius $0.2\,\text{m}$0.2m. The string does not slip on the pulley. Taking $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2: (a) explain why the two string tensions are unequal; (b) find the acceleration of the masses; (c) compare with the acceleration if the pulley were light.

Model solution. (a) A net torque is required to angularly accelerate the pulley ($\tau = I\alpha$τ=Iα with $I \ne 0$I=0); that torque is $(T_1 - T_2)R$(T1​−T2​)R, so $T_1 \ne T_2$T1​=T2​. With a light pulley $I = 0$I=0 and the tensions would be equal.

(b) Pulley: $I_p = \tfrac12(2)(0.2)^2 = \tfrac12(2)(0.04) = 0.04\ \text{kg m}^2$Ip​=21​(2)(0.2)2=21​(2)(0.04)=0.04 kg m2. Write the three equations (let $T_1$T1​ be on the $5\,\text{kg}$5kg side, $T_2$T2​ on the $3\,\text{kg}$3kg side, downward-positive for the heavy mass), using $\alpha = a/R$α=a/R:

$$\begin{aligned} 5g - T_1 &= 5a \quad (1)\\ T_2 - 3g &= 3a \quad (2)\\ (T_1 - T_2)R &= I_p\frac{a}{R} \ \Rightarrow\ T_1 - T_2 = \frac{I_p\,a}{R^2} = \frac{0.04\,a}{0.04} = a \quad (3) \end{aligned}$$5g−T1​T2​−3g(T1​−T2​)R​=5a(1)=3a(2)=Ip​Ra​ ⇒ T1​−T2​=R2Ip​a​=0.040.04a​=a(3)​

Add (1) and (2): $5g - 3g - (T_1 - T_2) = 8a$5g−3g−(T1​−T2​)=8a, i.e. $2g - a = 8a$2g−a=8a using (3). Hence

$$9a = 2g = 19.6 \ \Rightarrow\ a = \frac{19.6}{9} \approx 2.18\ \text{m s}^{-2}.$$9a=2g=19.6 ⇒ a=919.6​≈2.18 m s−2.

(c) A light pulley gives $a = \dfrac{(5-3)g}{5+3} = \dfrac{2g}{8} = 2.45\,\text{m s}^{-2}$a=5+3(5−3)g​=82g​=2.45m s−2. The massive pulley reduces the acceleration because part of the driving force now accelerates the pulley's own inertia. The mark-winning step is the substitution $\alpha = a/R$α=a/R turning equation (3) into a relation between tensions and $a$a, and the recognition that $I_p/R^2 = \tfrac12 M_p = 1\,\text{kg}$Ip​/R2=21​Mp​=1kg acts like an extra mass.

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5 Synoptic links