IB Physics HLRigid Body MechanicsHL onlyPaper 1 & 2~12 min read
Rotational Kinetic Energy
A spinning flywheel stores energy even when its centre isn’t going anywhere. A ball rolling down a hill arrives slower than one that just slides, because some of its energy has gone into spinning. Both come down to rotational kinetic energy — the energy a body carries purely because it’s turning. It completes the rotational toolkit, mirrors the linear kinetic energy you already know, and unlocks the elegant physics of rolling objects.
📘 What you need to know
A rotating body has rotational kinetic energy: Ek = ½Iω2
It can also be written in terms of angular momentum: Ek = L2 / 2I
These mirror the linear forms Ek = ½mv2 and Ek = p2 / 2m
Rolling without slipping combines rotation and translation; the contact point is momentarily at rest
A rolling object’s total kinetic energy is linear + rotational: Ek = ½mv2 + ½Iω2
Rolling down a slope: gravitational PE converts into both linear and rotational kinetic energy
The rotational kinetic energy formula
A body moving in a straight line has linear kinetic energy ½mv2. Swap mass for moment of inertia and speed for angular velocity, and you get the energy of a rotating body:
Rotational kinetic energyEk = ½Iω2Ek = L2 / 2I
where Ek is the rotational kinetic energy in joules (J), I is the moment of inertia in kg m2, ω is the angular velocity in rad s−1, and L is the angular momentum in kg m2 rad s−1. The two forms are exact partners of the linear ½mv2 and p2/2m — use whichever fits the quantities you’re given.
WE 1
A flywheel has a moment of inertia of 0.60 kg m2 and spins at 8.0 rad s−1. Calculate its rotational kinetic energy.
Step 1 — use Ek = ½Iω²
Ek = ½ × I × ω²
Step 2 — substituteEk = ½ × 0.60 × 8.0² = 0.30 × 64Ek = 19 J (2 s.f.)Same shape as ½mv² — rotational “mass” times rotational “speed” squared.
WE 2
A rotating body has an angular momentum of 12 kg m2 s−1 and a moment of inertia of 3.0 kg m2. Calculate its rotational kinetic energy using the angular-momentum form.
Step 1 — use Ek = L² ÷ 2I
Ek = L² ÷ (2I)
Step 2 — substituteEk = 12² ÷ (2 × 3.0) = 144 ÷ 6.0Ek = 24 JHandy when you know L and I but not ω — no need to find ω first.
Rolling without slipping
Wheels, balls and discs usually move by rolling — a combination of rotating and translating at the same time. When an object rolls without slipping, there’s just enough friction to stop it sliding, and this links its linear and angular speeds through v = ωr.
The striking consequence is that different points on the object move at different speeds. The point touching the ground is momentarily at rest (its rolling and sliding velocities cancel), the centre moves at v = ωr, and the top moves at 2v — twice the centre’s speed.
Rolling without slipping: the contact point is instantaneously at rest, the centre moves at v = ωr, and the top moves at 2v. Rotation and translation add together.
Because a rolling object is doing both things at once, its total kinetic energy is the sum of a linear part and a rotational part:
Total kinetic energy of a rolling objectEk total = ½mv2 + ½Iω2
Linear KE ½mv²
+
Rotational KE ½Iω²
=
Total KE rolling
Rolling down a slope
When an object rolls down a slope, its gravitational potential energy at the top converts into both linear and rotational kinetic energy at the bottom. This is why a rolling ball is always slower than a frictionless sliding one from the same height — some of the energy has to go into making it spin.
Rolling down a slope: gravitational PE (mgΔh) at the top becomes linear and rotational kinetic energy at the bottom. The ball arrives slower than a frictionless slider would.
For a solid sphere (moment of inertia ⅖mr2), substituting v = ωr and equating the energies gives a compact result for the total energy at the bottom:
Solid sphere rolling from height ΔhmgΔh = 7/10 mω2r2
WE 3
A solid sphere (moment of inertia ⅖mr2) rolls without slipping from rest down a slope, dropping a vertical height of 2.0 m. Calculate its linear speed at the bottom. Take g = 9.81 m s−2.
Step 1 — energy conservation: mgΔh = ½mv² + ½Iω²
mgΔh = ½mv² + ½(⅖mr²)(v/r)²
Step 2 — simplify (the m and r cancel through)mgΔh = ½mv² + ⅕mv² = 7/10 mv²Step 3 — rearrange for v and substitutev = √(10gΔh ÷ 7) = √(10 × 9.81 × 2.0 ÷ 7)v = 5.3 m s⁻¹ (2 s.f.)A frictionless slide would reach 6.3 m s⁻¹ — the rolling sphere is slower because energy went into spin.
Here’s the intuition for why a rolling ball loses a race to a sliding block: the drop gives them the same energy budget, mgΔh. The slider spends all of it on going forwards, but the roller has to split its budget between moving and spinning. Less energy for forward motion means a lower final speed. And the more spread out the object’s mass (bigger I), the bigger the rotational share — so a hoop rolls down slower than a solid sphere.
WE 4
A flywheel of mass M and radius R (moment of inertia ½MR2) spins at angular velocity ω with rotational kinetic energy EK. A second flywheel of mass ½M and radius ½R is dropped on top, and the combination now spins at ⅔ω. Show that the new kinetic energy is EK / 2.
Step 1 — original energyEK = ½(½MR²)ω² = ¼MR²ω²Step 2 — combined moment of inertiaInew = ½MR² + ½(½M)(½R)² = 9/16 MR²Step 3 — new energy at ωnew = ⅔ωE = ½(9/16 MR²)(⅔ω)² = ½(¼MR²ω²)E = ½EKEnergy isn’t conserved here — dropping the second wheel on is like an inelastic (sticking) collision.
🛠️ Solving a rotational energy problem
Spinning only? Use Ek = ½Iω2 (or L2/2I if you have L and I).
Rolling? Add both parts: Ek = ½mv2 + ½Iω2, and link them with v = ωr.
Down a slope? Set mgΔh = total kinetic energy at the bottom.
Substitute the moment of inertia and v = ωr, then simplify (mass and radius often cancel).
Solve for the speed or angular velocity you need.
💡 Top tips
Two forms, one energy.Ek = ½Iω2 = L2/2I — pick the one matching your data.
Rolling is two energies. Always add the linear and rotational parts for a rolling object.
Link with v = ωr. This connects the linear and rotational speeds so terms combine.
Mass and radius often cancel on a slope — the final speed usually doesn’t depend on them.
Quick recap: A rotating body has rotational kinetic energy Ek = ½Iω2 = L2/2I, mirroring the linear forms. A rolling object has both linear and rotational kinetic energy, linked by v = ωr, with its contact point momentarily at rest. Rolling down a slope, gravitational PE splits into both kinds, so a rolling object arrives slower than a frictionless slider.
⚠ Common mistakes
Forgetting the rotational part — a rolling object has both ½mv2 and ½Iω2
Not linking v and ω with v = ωr before combining terms
Assuming a rolling ball reaches the same speed as a sliding one from the same height
Using the wrong moment of inertia for the shape (sphere vs hoop vs disc)
Thinking energy is conserved when two spinning objects join together (it usually isn’t)
That completes the whole Rigid Body Mechanics topic — from torque and rotational equilibrium, through the angular kinematics and dynamics, to angular momentum, impulse and now rotational energy. You’ve built the full rotational mirror of linear mechanics. Brilliant work getting through it — every rotating system in the course now sits on the foundations you’ve just laid.
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