Short on time? This page is the recap. For the derivation, diagrams and worked examples, use the full study guide. This is the algebra-based course: no calculus is needed.
Recap
- Every piece of a spinning rigid system moves at v = rω, so every piece has ordinary kinetic energy ½mv².
- Adding those energies gives K = ½Iω², where I = Σmr². For a fixed axis, this is the object’s whole kinetic energy, not an extra amount.
- A system that moves and spins has translational kinetic energy (centre of mass) plus rotational kinetic energy (spin about the centre of mass).
- A flywheel on a fixed axle has kinetic energy even though its centre of mass is at rest and its total momentum is zero.
- Rotational kinetic energy is a scalar. Spinning clockwise or counterclockwise makes no difference to K.
Key relationships
| Relationship | Symbols and units | Use it to |
|---|---|---|
| v = rω | v in m/s, r in m, ω in rad/s | find the speed of a point at distance r |
| I = Σmr² | kg·m² | rotational inertia of a few point objects |
| K = ½Iω² | J | rotational kinetic energy about a given axis |
| K_total = ½Mv_cm² + ½I_cm ω² | J | an object that moves and spins |
| I = I_cm + Md² | kg·m² | link the pivot and centre-of-mass descriptions |
| ω = rev/min × 2π ÷ 60 | rad/s | convert machine speeds |
Factors of change: ω × 2 → K × 4; ω × 3 → K × 9; I × 2 → K × 2. A hoop (MR²) stores twice the energy of a disc (½MR²) of the same mass and radius at the same ω.
Assumptions behind the numbers
- The system is rigid: every part has the same ω and keeps its distance from the axis.
- Small objects are treated as points, so I = mr² for each one.
- Rotational inertias of extended objects (hoops, discs, rods) are given.
- Rotation happens in one plane about one axis.
Mistakes to avoid
- Using rev/min or degrees per second. Convert to rad/s first.
- Double counting. About a fixed pivot use ½I_pivot ω² alone; otherwise use ½Mv_cm² + ½I_cm ω². Never mix the two.
- “No centre-of-mass motion, so no kinetic energy.” Spinning objects have kinetic energy.
- Making K negative for clockwise rotation.
- Treating K as proportional to ω. It is proportional to ω².
- Ignoring mass distribution. Same mass, same ω, different I gives different K.
Quick self-check
- A wheel with I = 0.40 kg·m² spins at 5.0 rad/s. What is its kinetic energy? (5.0 J)
- Its angular velocity is halved. By what factor does K change? (× ¼)
- A 2.0 kg ball moves at 3.0 m/s and spins at 30 rad/s, with I_cm = 0.010 kg·m². What is its total kinetic energy? (9.0 J + 4.5 J = 13.5 J)
Next: practice questions.