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5b Energy methods for rolling and rotating bodies

A rolling object down a ramp, or a mass falling on a string attached to a flywheel, recurs every year — the skill is splitting the kinetic energy correctly between translation and rotation.

The energy-conservation chain for a rolling object

  • Gravitational PE lost = total kinetic energy gained: mgh = ½mv² + ½Iω².
  • For ROLLING WITHOUT SLIPPING, the translational and rotational speeds are linked: v = rω, so ω = v/r — substitute this before solving for the unknown.
  • The MOMENT OF INERTIA form matters: a solid sphere (I = ⅖mr²), a solid disc/cylinder (I = ½mr²), and a hoop/ring (I = mr²) all reach the bottom of the same ramp at different speeds, because a different fraction of their KE goes into rotation.
  • An object with a SMALLER I (relative to mr²) converts less of its PE into rotational KE, leaving more for translational KE — so it reaches the bottom FASTER.
Real paper — 2023 Finding rotational inertia from a rolling barrel An empty barrel (m = 5.50 kg, r = 0.280 m) rolls down a 1.34 m ramp, reaching 3.40 m s⁻¹ at the bottom. Find its rotational inertia. GPE lost: mgh = 5.50 × 9.81 × 1.34 = 72.3 J. This equals total KE: ½mv² + ½Iω², with ω = v/r = 3.40/0.280 = 12.1 rad s⁻¹. Translational KE: ½ × 5.50 × 3.40² = 31.8 J. Rotational KE = total − translational = 72.3 − 31.8 = 40.5 J, so ½Iω² = 40.5 → I = 2 × 40.5 / 12.1² = 0.553 kg m². ✓ I ≈ 0.553 kg m² — found entirely from energy conservation, without ever needing the barrel's actual mass distribution. ✕

Forgetting the rotational KE term entirely

A rolling (not sliding) object always carries BOTH translational KE (½mv²) AND rotational KE (½Iω²) — using mgh = ½mv² alone, as if the object were simply sliding, overstates its final linear speed and is one of the most common errors on this topic. Always ask "is this object rolling?" before choosing which energy terms to include. Real paper — 2024 How a different moment of inertia changes the final speed A ball and cup fall 0.750 m, driving a flywheel. Explain how a solid disc of the same mass and radius (instead of the flywheel's actual shape) changes the final velocity of the ball. A flywheel is built with most of its mass out near the rim. A solid disc of the same mass and radius spreads it evenly, so on average its mass sits closer to the axle. Since I ∝ mr², the disc has the SMALLER rotational inertia. The same GPE is released as the ball and cup fall 0.750 m: mgh = ½mv² + ½Iω², with ω = v/r. A smaller I takes less of that energy into rotation, leaving more for the ball and cup. Forces agree: with a smaller I the wheel spins up more easily (τ = Iα), so the cup accelerates faster (a = rα) and is moving faster after the same drop. The tension falls a little as the acceleration rises, but the acceleration still goes up: mg = a(m + I/r²). ✓ The ball and cup end up moving FASTER. The disc keeps its mass closer to the axle, so its I is smaller than the flywheel's; less of the released energy goes into turning it and more into the falling ball and cup. The full worked solution is 2024 Q2(d) under What actually comes up.

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