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Lesson 0 built the rotational analogue of Newton's second law, τ = Iα. This lesson completes the linear-to-rotational dictionary by introducing angular momentum L = Iω as the rotational counterpart of linear momentum p = mv, and rotational kinetic energy ½Iω² as the counterpart of ½mv². It also closes the conservation-law picture: in a closed system, just as linear momentum is conserved when no external force acts, angular momentum is conserved when no external torque acts. The conservation principle is the engineering payoff. It explains why a figure skater spins faster when pulling in her arms, why a diver tucks to rotate faster mid-flight, why a satellite's spin rate increases when its solar panels retract, and why a flywheel mounted on a frictionless bearing maintains its angular velocity indefinitely. Power delivery to a rotating shaft is P = τω; rolling-motion problems require both the translational kinetic energy of the centre of mass and the rotational kinetic energy about the centre of mass, with the no-slip condition v = rω linking the two.
Spec mapping (AQA 7408 §3.11.1 — Rotational dynamics, option C, continued): This lesson covers angular momentum L = Iω, conservation of angular momentum in a closed system, rotational kinetic energy E_rot = ½Iω², work done by a torque W = τΔθ, power delivered as P = τω, the full linear–rotational analogy table, and rolling motion as combined translation plus rotation with the no-slip constraint v = rω. (Refer to the official AQA specification document for exact wording.)
Synoptic links: Conservation of angular momentum is the rotational sibling of Year 12 conservation of linear momentum (§3.4 mechanics) and is structurally analogous to conservation of charge and conservation of energy. It is examined synoptically with circular motion (§3.6.1) where the centripetal acceleration analysis assumes constant ω — but the value of that ω is set by the angular-momentum history of the system. It also feeds directly into the engine-cycle work calculations of §3.11.2: the work done in one revolution of a crankshaft is τ̄ × 2π, and the mean engine power is τ̄ ω. In particle physics, the same algebra describes electron spin angular momentum (with the additional twist that quantum-mechanical L is quantised in units of ħ — out of A-Level scope but a useful pointer to Cambridge interview-level extensions).
For a rigid body rotating about a fixed axis with angular velocity ω and moment of inertia I about that axis:
Angular momentum: L = Iω, measured in kg m² s⁻¹ (or equivalently N m s).
For a point particle of mass m moving with linear velocity v at perpendicular distance r from a reference axis, the angular momentum about that axis is L = mvr (the cross product r × p in vector language).
The rotational analogue of Newton's second law in momentum form, F = dp/dt, is:
τ_net = dL/dt
If the net external torque on a system is zero, dL/dt = 0 and angular momentum is conserved:
Conservation of angular momentum: In an isolated system (no external torques), the total angular momentum L_total = Σ Iᵢ ωᵢ is constant.
This is the rotational sibling of conservation of linear momentum. It applies even when internal forces redistribute mass within the system — those internal forces produce equal-and-opposite torques that cancel pairwise.
A figure skater spinning with arms outstretched has a relatively large moment of inertia I₁ and a corresponding angular velocity ω₁. When she pulls her arms in, her mass distribution becomes more concentrated about the spin axis and her moment of inertia drops to I₂ < I₁. Because there is no external torque (ice is approximately frictionless), L is conserved:
I₁ ω₁ = I₂ ω₂
Since I₂ < I₁, it must be that ω₂ > ω₁ — she spins faster. The angular velocity ratio is the inverse of the moment-of-inertia ratio: ω₂/ω₁ = I₁/I₂.
The same principle underlies a diver's somersault: arms-and-legs-extended gives a larger I and a slower rotation off the springboard; tucking pulls mass close to the spin axis, dropping I and increasing ω. To enter the water cleanly the diver extends again, raising I and dropping ω back to a low value just before entry.
Question: A skater is spinning at 2.0 rev s⁻¹ with arms extended, with a moment of inertia of 5.0 kg m². She pulls her arms in, reducing her moment of inertia to 2.0 kg m². Calculate (a) her new angular velocity in rev s⁻¹, (b) the ratio of her new rotational kinetic energy to the original.
Solution:
(a) Conserve angular momentum: I₁ ω₁ = I₂ ω₂. ω₂ = ω₁ (I₁/I₂) = 2.0 × (5.0/2.0) = 5.0 rev s⁻¹.
(b) Rotational KE is ½Iω². The ratio is: KE₂/KE₁ = (I₂ ω₂²) / (I₁ ω₁²) = (2.0 × 5.0²) / (5.0 × 2.0²) = 50/20 = 2.5.
Her rotational kinetic energy increases by a factor of 2.5. The extra energy came from the work she did against the centrifugal pull when pulling her arms inward — chemical energy in her muscles converted to rotational KE. Angular momentum is conserved; rotational KE is not (because internal work was done).
Exam Tip: Conservation of angular momentum does not imply conservation of rotational kinetic energy. When I changes through internal action, KE_rot = L²/(2I) — so as I drops, KE_rot rises. Examiners frequently test this distinction.
A rotating body of moment of inertia I and angular velocity ω stores kinetic energy:
E_rot = ½ I ω²
This is the direct analogue of ½mv² for translation. It is derived by integrating over all the mass elements: each element at distance rᵢ moves with linear speed vᵢ = rᵢω, contributing ½mᵢvᵢ² = ½mᵢrᵢ²ω². Summing over the body gives E = ½ ω² Σ mᵢ rᵢ² = ½ I ω².
Question: A solid steel flywheel modelled as a uniform disc has mass 200 kg and radius 0.50 m. It spins at 3000 rpm. Calculate the rotational kinetic energy stored.
Solution:
I = ½ M R² = 0.5 × 200 × 0.50² = 25.0 kg m². ω = 3000 × 2π / 60 = 100π ≈ 314.2 rad s⁻¹. E_rot = ½ × 25.0 × 314.2² = 12.5 × 98 696 ≈ 1.23 × 10⁶ J (about 1.2 MJ).
That's roughly the kinetic energy of a 1-tonne car at 50 m s⁻¹. Industrial flywheels for grid-stabilisation can store many tens of megajoules, with the energy released in a few seconds when needed.
The work done by a constant torque τ during an angular displacement Δθ is:
W = τ Δθ
This is the direct analogue of W = F Δx for linear work. For a variable torque, integrate: W = ∫ τ dθ.
The power delivered by a torque rotating at angular velocity ω is the time derivative of work:
P = τ ω
This is the rotational analogue of P = Fv. A motor producing a torque of 50 N m driving a shaft at 100 rad s⁻¹ delivers 5000 W = 5 kW. The same motor at half the speed delivers half the power even though the torque is unchanged. This is why an internal-combustion engine's power and torque curves are different shapes — power = torque × ω, so peak power typically occurs at a higher ω than peak torque.
Question: A drilling motor produces a constant torque of 8.0 N m and runs at 1500 rpm. Calculate (a) the power delivered, (b) the work done in 30 s of continuous drilling.
Solution:
(a) ω = 1500 × 2π / 60 = 50π ≈ 157.1 rad s⁻¹. P = τ ω = 8.0 × 157.1 = 1.26 × 10³ W ≈ 1.26 kW.
(b) W = P × t = 1257 × 30 = 3.77 × 10⁴ J ≈ 37.7 kJ. (Equivalent to Δθ = ω × t = 157.1 × 30 = 4712 rad, and W = τΔθ = 8.0 × 4712 = 3.77 × 10⁴ J. ✓)
The full dictionary between linear and rotational mechanics:
| Quantity | Linear | Rotational |
|---|---|---|
| Displacement | x (m) | θ (rad) |
| Velocity | v (m s⁻¹) | ω (rad s⁻¹) |
| Acceleration | a (m s⁻²) | α (rad s⁻²) |
| Inertia | mass m (kg) | moment of inertia I (kg m²) |
| Force / cause of acceleration | F (N) | torque τ (N m) |
| Newton's second law | F = ma | τ = Iα |
| Momentum | p = mv | L = Iω |
| Newton-II in momentum form | F = dp/dt | τ = dL/dt |
| Kinetic energy | ½ m v² | ½ I ω² |
| Work | W = F Δx | W = τ Δθ |
| Power | P = F v | P = τ ω |
| Conservation (closed system) | Σp constant | ΣL constant |
Notice how every rotational equation is structurally identical to its linear sibling. Once you master one column, the other is essentially a rebrand.
A wheel or ball that rolls (rather than slides) has both translational motion of its centre of mass and rotation about that centre. The total kinetic energy is the sum:
KE_total = ½ M v² + ½ I ω²
where v is the speed of the centre of mass and I is the moment of inertia about the centre-of-mass axis.
For rolling without slipping, the contact point is momentarily at rest, which imposes the constraint:
v = R ω (rolling-without-slipping condition)
This is the most important kinematic constraint in rolling problems. It lets you eliminate either v or ω from the energy equation.
For example, a solid uniform sphere (I = ⅖MR²) rolling without slipping has:
KE = ½ M v² + ½ × (2/5) M R² × ω² = ½ M v² + ½ × (2/5) M v² = ½ M v² (1 + 2/5) = (7/10) M v²
So 5/7 of the energy is translational and 2/7 is rotational. For a solid disc/cylinder (I = ½MR²), the ratio is 2:1 translation:rotation. For a hoop (I = MR²), it's 1:1 — a hoop dedicates half its kinetic energy to spinning, which is why hoops roll downhill more slowly than discs of equal mass and radius.
Question: A solid uniform sphere, a solid uniform disc and a thin hoop, all of equal mass M and equal radius R, are released from rest at the top of a frictionless ramp of height h and roll without slipping to the bottom. Which reaches the bottom first?
Solution:
Using energy conservation, Mgh = ½Mv² + ½Iω², and applying v = Rω, the final speed of each is:
Sphere: Mgh = ½Mv² + ½(⅖MR²)(v²/R²) = (7/10)Mv² ⇒ v_sphere = √(10gh/7).
Disc: Mgh = ½Mv² + ½(½MR²)(v²/R²) = (3/4)Mv² ⇒ v_disc = √(4gh/3).
Hoop: Mgh = ½Mv² + ½(MR²)(v²/R²) = Mv² ⇒ v_hoop = √(gh).
Ratios: v_sphere : v_disc : v_hoop = √(10/7) : √(4/3) : 1 ≈ 1.195 : 1.155 : 1.000.
The sphere wins, the disc is second, the hoop is last. The order depends only on the fractional pre-factor in the moment of inertia formula — not on mass, not on radius. The greater the fraction of mass concentrated near the axis (smaller fraction in front of MR²), the faster the body accelerates down the ramp because less of the released gravitational PE is locked up in rotational KE.
Exam Tip: Rolling-race problems are a perennial favourite at A-Level interview and at A2 paper level. The mass and radius cancel — a hoop the size of a planet rolls at the same final speed as a hoop the size of a coin from the same height. The moment-of-inertia fraction is everything.
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