Short on time? This page is the recap for the calculus-based course (separate from Physics 1, which does not integrate over continuous bodies). For the derivations, graphs and worked examples, use the full study guide.
Recap
- Every piece of a rigid body shares one ω. A piece at distance r from the axis moves at v = rω (ω in rad/s).
- Adding ½v² dm over the body gives K = ½ω²∫r² dm = ½Iω². For a fixed axis this is the total kinetic energy, not an extra amount.
- For a body that moves and spins, the total splits into translation of the centre of mass plus rotation about it. The cross term vanishes because ∫v′ dm = 0 in the centre-of-mass frame.
- A body can have kinetic energy with its centre of mass at rest (a flywheel on a fixed axle). Its momentum is zero; its kinetic energy is not.
- Rotational kinetic energy is a scalar and never negative. Opposite spins add.
Key relationships
| Relationship | When to use it | Note |
|---|---|---|
| K = ∫½v² dm | Always | Start here for non-uniform bodies |
| K = ½Iω² | Fixed axis | I about that axis |
| K = ½Mv_cm² + ½I_cm ω² | Any rigid motion | I about the centre of mass |
| I_pivot = I_cm + Md² | Linking the two | Shows both methods agree |
| ω (rad/s) = rev/min × 2π/60 | Before any calculation | Unit: kg·m²·rad²/s² = J |
| Constant α from rest: K = ½Iα²t² = Iαθ | Graphs | K–t parabola; K–θ straight line |
Assumptions behind the numbers
- The body is rigid: every piece has the same ω.
- I is about the stated axis; for the split formula, about a parallel axis through the centre of mass.
- Inertial reference frame; air resistance ignored unless stated.
Mistakes to avoid
- Double counting: ½I_pivot ω² + ½Mv_cm² is wrong.
- Tip or rim speed in ½Mv². Pieces move at different speeds.
- rev/min or degrees in K = ½Iω².
- “Centre of mass at rest, so K = 0.”
- Giving K a sign or letting opposite spins cancel.
- Uniform-rod formulas for a non-uniform rod. Integrate with λ(x).
- “Larger I always means larger K.” Only at the same ω.
Quick self-check
- I = 0.20 kg·m², ω = 5.0 rad/s. Find K. (½ × 0.20 × 25 = 2.5 J)
- M = 2.0 kg, v_cm = 3.0 m/s, I_cm = 0.10 kg·m², ω = 10 rad/s. Find the total K. (9.0 J + 5.0 J = 14 J)
- A flywheel with I = 0.050 kg·m² spins at 600 rev/min. Find K. (ω = 62.8 rad/s; K = 98.7 J ≈ 99 J)
Next: practice questions.