Angular Momentum and Conservation

Angular momentum is the rotational counterpart of linear momentum, and its conservation is one of the deepest and most useful principles in mechanics. Where linear momentum $p = mv$ is conserved when no external force acts, angular momentum $L = I\omega$ is conserved when no external torque acts — and the two conservation laws are independent, springing from the symmetry of space under translation and rotation respectively. This lesson defines $L$ for both rotating bodies and free particles, establishes $\tau = \mathrm{d}L/\mathrm{d}t$ , and then deploys conservation in the situations Paper 3 loves: the spinning skater who changes shape, the dropped ring, the ballistic pendulum where a particle strikes and sticks to a pivoted rod. A recurring theme — and a frequent examiner trap — is that angular momentum can be conserved while kinetic energy is not.

1 Where this sits in AQA 7367

This is Paper 3, the Mechanics option (7367/3M), the natural sequel to rotational dynamics. Conservation arguments are heavily AO2/AO3: you must recognise that angular momentum is conserved (and about which axis), justify it from the absence of external torque, and then solve a multi-stage problem (often a collision with energy loss). Paper 3 weights these objectives AO1 40% / AO2 25% / AO3 35%. The prerequisite is the moment-of-inertia toolkit and the linear momentum/impulse ideas of A-Level Mechanics, of which this is the rotational mirror.

2 Core theory — defining angular momentum

For a rigid body rotating about a fixed axis with angular speed $\omega$ and moment of inertia $I$ about that axis,

\boxed{\,L = I\omega\,}.

For a single particle of mass $m$ moving with speed $v$ , the angular momentum about a point $O$ is

L = mvd,

where $d$ is the perpendicular distance from $O$ to the line of motion. (These agree: a particle moving in a circle of radius $r$ has $L = mvr = m(r\omega)r = mr^2\omega = I\omega$ , since $I = mr^2$ for a point mass.)

Property	Detail
SI unit	$\text{kg m}^2\,\text{s}^{-1}$ (equivalently $\text{N m s}$ )
Vector	directed along the rotation axis by the right-hand rule
Depends on the reference point/axis	$d$ (or $I$ ) is measured from that axis

The link to torque

Differentiating $L = I\omega$ (constant $I$ ) gives the rotational form of Newton's second law in momentum language:

\boxed{\,\tau = \frac{\mathrm{d}L}{\mathrm{d}t}\,}.

This is the exact mirror of $F = \dfrac{\mathrm{d}p}{\mathrm{d}t}$ . Two immediate consequences:

if $I$ is constant, $\tau = I\dfrac{\mathrm{d}\omega}{\mathrm{d}t} = I\alpha$ (last lesson's law);
if the external torque is zero, $\dfrac{\mathrm{d}L}{\mathrm{d}t} = 0$ , so $L$ is constant — conservation of angular momentum.

Conservation of angular momentum

When no external torque acts about the chosen axis, the total angular momentum is conserved. If a body's moment of inertia changes from $I_1$ to $I_2$ (it changes shape, or merges with another body),

\boxed{\,I_1\omega_1 = I_2\omega_2\,}.

Pull mass inwards and $I$ falls, so $\omega$ rises — the skater spins faster. The striking fact is that kinetic energy is generally not conserved in these processes: $\text{KE} = \tfrac12 I\omega^2 = \tfrac{(I\omega)^2}{2I} = \dfrac{L^2}{2I}$ , so at fixed $L$ , decreasing $I$ increases KE (the skater does internal work pulling her arms in), while merging bodies (increasing $I$ ) loses KE to heat. Angular momentum and kinetic energy are different bookkeeping, and only one is conserved when shape changes.

Angular impulse

Integrating $\tau = \mathrm{d}L/\mathrm{d}t$ over time gives the angular impulse–momentum theorem:

\int \tau\,\mathrm{d}t = \Delta L = I\omega_2 - I\omega_1, \qquad (\text{constant }\tau:\ \tau\,\Delta t = \Delta L).

Angular momentum of a single particle in the plane

For a particle at position $\mathbf{r} = (x,y)$ moving with velocity $\mathbf{v} = (v_x, v_y)$ , the angular momentum about the origin is the $z$ -component of $m\,\mathbf{r}\times\mathbf{v}$ :

L = m(x v_y - y v_x).

For a system of particles these add: $L = \sum_i m_i(x_i v_{yi} - y_i v_{xi})$ . When the particle moves in a straight line, $L = mvd$ is constant (the perpendicular distance $d$ to the line never changes), which is consistent with $\tau = 0$ for a free particle — a useful reassurance that the two definitions agree.

A point that repays attention: the value of $L$ depends on the choice of reference point. The same particle has different angular momenta about different points, so in any problem you must name the axis before computing — and you must use the same axis on both sides of a conservation equation. The natural choice in collision problems is the pivot or the centre of mass, because the awkward impulsive forces act through that point and so contribute no torque.

3 Worked examples with mark scheme

Example 1 — the spinning skater

A skater has moment of inertia $4\,\text{kg m}^2$ spinning at $2\,\text{rev s}^{-1}$ with arms extended. She pulls them in, reducing $I$ to $1.5\,\text{kg m}^2$ . Find her new rate of spin, and the change in kinetic energy.

No external torque about the vertical axis, so $L$ is conserved: (M1)

I_1\omega_1 = I_2\omega_2 \ \Rightarrow\ 4(2) = 1.5\,\omega_2 \ \Rightarrow\ \omega_2 = \frac{8}{1.5} \approx 5.33\ \text{rev s}^{-1}. \quad (\textbf{M1 A1})

Working in $\text{rad s}^{-1}$ : $\omega_1 = 2(2\pi) = 4\pi$ , $\omega_2 = \dfrac{8}{1.5}(2\pi) = \dfrac{32\pi}{3}$ . Then

\text{KE}_1 = \tfrac12(4)(4\pi)^2 = 32\pi^2 \approx 316\ \text{J}, \quad \text{KE}_2 = \tfrac12(1.5)\!\left(\tfrac{32\pi}{3}\right)^2 = \tfrac{256}{3}\pi^2 \approx 842\ \text{J}. \quad (\textbf{M1})

The KE increases by about $526\,\text{J}$ ; the skater supplies this by doing work pulling her arms in against the outward "centrifugal" tendency. (A1)

(M1 conservation stated; M1 solve for $\omega_2$ ; A1 $5.33\,\text{rev s}^{-1}$ ; M1 both KEs in consistent units; A1 increase identified and explained. The same units must be used throughout — mixing rev and rad is the classic error.)

Example 2 — the dropped ring (a rotational "inelastic collision")

A disc, $I = 0.5\,\text{kg m}^2$ , rotates at $10\,\text{rad s}^{-1}$ . A ring, $I = 0.3\,\text{kg m}^2$ , is dropped concentrically onto it and they rotate together. Find the common angular speed and the kinetic energy lost.

Dropping the ring exerts no torque about the spin axis, so $L$ is conserved: (M1)

I_1\omega_1 = (I_1 + I_2)\omega \ \Rightarrow\ 0.5(10) = 0.8\,\omega \ \Rightarrow\ \omega = 6.25\ \text{rad s}^{-1}. \quad (\textbf{M1 A1})

\Delta\text{KE} = \tfrac12(0.5)(10)^2 - \tfrac12(0.8)(6.25)^2 = 25 - 15.625 = 9.375\ \text{J lost}. \quad (\textbf{M1 A1})

(M1 conservation; M1 solve; A1 $\omega = 6.25$ ; M1 both KEs; A1 $9.375\,\text{J}$ . The energy goes to friction as the surfaces reach a common speed — the rotational analogue of a perfectly inelastic collision.)

Example 3 — angular impulse

A constant torque of $8\,\text{N m}$ acts on a flywheel $(I = 2\,\text{kg m}^2)$ , initially at rest, for $3\,\text{s}$ . Find the final angular speed.

\tau\,\Delta t = I\omega - 0 \ \Rightarrow\ 8(3) = 2\omega \ \Rightarrow\ \omega = 12\ \text{rad s}^{-1}. \quad (\textbf{M1 A1})

Example 4 — a bead sliding on a rotating wire

A small bead of mass $m$ is threaded on a smooth straight wire that rotates freely in a horizontal plane about a fixed vertical axis through one end. Initially the wire turns at $\omega_0$ with the bead at distance $r_0$ from the axis. The bead is released and slides outward; find its angular speed when it reaches distance $r$ , and comment on the wire's tendency to slow.

The only horizontal force on the bead from the smooth wire is perpendicular to the wire (the wire cannot push along its own length), and that force passes through... no — it acts on the bead at distance $r$ , so it does exert a torque on the bead; but by Newton's third law the bead exerts an equal and opposite torque on the wire. Treating the bead-plus-wire as the system with no external torque about the axis (the wire is "freely" mounted), angular momentum is conserved. Modelling the bead as a point mass with $I = mr^2$ and neglecting the wire's own inertia: (M1)

I_0\omega_0 = I\omega \ \Rightarrow\ mr_0^2\,\omega_0 = mr^2\,\omega \ \Rightarrow\ \omega = \omega_0\frac{r_0^2}{r^2}. \quad (\textbf{M1 A1})

As the bead slides outward ( $r > r_0$ ), $\omega$ decreases — the system spins slower as mass moves outward, the exact reverse of the skater pulling in. The kinetic energy of rotation $\tfrac12 I\omega^2 = \dfrac{L^2}{2I}$ falls as $I$ grows, the deficit supplied as the bead's outward kinetic energy along the wire. (A1)

(M1 conservation of $L$ for the bead-wire system; M1 $I = mr^2$ substituted; A1 $\omega = \omega_0 r_0^2/r^2$ ; A1 correct qualitative comment. This is the everyday "merry-go-round" effect: walk outward and the roundabout slows.)

4 Specimen-style exam question — ballistic pendulum (particle strikes a rod)

(specimen-style — not from any past paper)

A uniform rod of mass $4\,\text{kg}$ and length $1\,\text{m}$ is freely pivoted at one end and hangs vertically at rest. A particle of mass $0.5\,\text{kg}$ , moving horizontally at $20\,\text{m s}^{-1}$ , strikes the free (lower) end and embeds itself. (a) Find the angular speed of the rod-and-particle immediately after impact. (b) Find the fraction of kinetic energy lost in the impact. (c) Find the maximum angle the rod swings through after impact.

Model solution. (a) Take angular momentum about the pivot, because the (large, impulsive) reaction at the pivot exerts no torque about the pivot. Before impact only the particle moves: $L = mvd = 0.5(20)(1) = 10\ \text{kg m}^2\,\text{s}^{-1}$ . After impact the system rotates with

I_{\text{total}} = I_{\text{rod}} + I_{\text{particle}} = \tfrac13(4)(1)^2 + 0.5(1)^2 = \tfrac43 + \tfrac12 = \tfrac{11}{6}\ \text{kg m}^2.

Conserving angular momentum about the pivot: $10 = \tfrac{11}{6}\,\omega \Rightarrow \omega = \dfrac{60}{11} \approx 5.45\ \text{rad s}^{-1}$ .

(b) Before: $\text{KE} = \tfrac12(0.5)(20)^2 = 100\,\text{J}$ . After: $\text{KE} = \tfrac12 I_{\text{total}}\omega^2 = \tfrac12\cdot\tfrac{11}{6}\cdot\left(\tfrac{60}{11}\right)^2 = \tfrac12\cdot\tfrac{11}{6}\cdot\tfrac{3600}{121} = \dfrac{300}{11} \approx 27.3\,\text{J}$ . Fraction lost $= 1 - \dfrac{300/11}{100} = 1 - \dfrac{3}{11} = \dfrac{8}{11} \approx 0.727$ , i.e. about $73\%$ is lost to the impact.

(c) After impact mechanical energy is conserved (the impulsive loss is over). Measuring the swing angle $\phi$ from the downward vertical, the centre of mass of the rod rises by $0.5(1-\cos\phi)$ and the particle by $1(1-\cos\phi)$ . Energy balance (all the post-impact KE becomes PE at the highest point):

\tfrac12 I_{\text{total}}\omega^2 = \bigl[m_{\text{rod}}g(0.5) + m_{\text{p}}g(1)\bigr](1-\cos\phi) = \bigl(4(9.8)(0.5) + 0.5(9.8)(1)\bigr)(1-\cos\phi) = 24.5(1-\cos\phi).

With the post-impact KE $= \tfrac{300}{11} \approx 27.27\,\text{J}$ :

27.27 = 24.5(1-\cos\phi) \ \Rightarrow\ 1 - \cos\phi = 1.113 \ \Rightarrow\ \cos\phi = -0.113 \ \Rightarrow\ \phi \approx 96.5°.

So the rod swings to about $96.5°$ , just past the horizontal, and momentarily stops. (To go right over the top would need $24.5(1-\cos 180°) = 49\,\text{J}$ , more than the $27.27\,\text{J}$ available, so it does not.) The mark-winning move throughout is taking moments about the pivot so the unknown impulsive reaction there drops out.

Angular Momentum and Conservation

Angular Momentum and Conservation

1 Where this sits in AQA 7367

2 Core theory — defining angular momentum

The link to torque

Conservation of angular momentum

Angular impulse

Angular momentum of a single particle in the plane

3 Worked examples with mark scheme

Example 1 — the spinning skater

Example 2 — the dropped ring (a rotational "inelastic collision")

Example 3 — angular impulse

Example 4 — a bead sliding on a rotating wire

4 Specimen-style exam question — ballistic pendulum (particle strikes a rod)

5 Synoptic links

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