Rotational Energy & Work
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
A spinning object has kinetic energy K = ½Iω². A torque turning through an angle does work W = ∫τ dθ, and the rotational work–energy theorem says Wnet = ΔKrot. Energy conservation works for rotation just as it does for sliding.
1. Rotational energy
2. Using it
Angles in W = τΔθ must be in radians: 3 revolutions is 6π rad.
For an object rotating about a fixed pivot, K = ½Ipivotω² covers all of its kinetic energy. For rolling objects, it’s easier to split it: ½Mvcm² + ½Icmω².
At the same ω, an object with its mass farther out stores more energy. A hoop has twice the kinetic energy of a disk with the same mass, radius and ω.
3. Example, then try it
A 2.0 kg uniform disk of radius 0.30 m spins at 20 rad/s. Find its kinetic energy. How much work does a constant 8.0 N·m torque do turning it through 5 revolutions?
I = ½(2.0)(0.30)² = 0.090 kg·m², so K = ½(0.090)(20)² = 18 J.
W = τΔθ = (8.0)(5 × 2π) ≈ 250 J.
A uniform rod 1.2 m long is pivoted at one end and released from rest while horizontal (the figure below). Find its angular velocity and the speed of its free end as it swings through vertical. (g = 10 m/s²)
Energy: the center of mass drops L/2, so Mg(L/2) = ½(ML²/3)ω².
Solve: ω² = 3g/L = 3(10) ÷ 1.2 = 25, so ω = 5.0 rad/s. The mass cancels.
Free end: v = ωL = (5.0)(1.2) = 6.0 m/s, faster than a ball dropped 1.2 m (√(2gL) ≈ 4.9 m/s).
Answer: 5.0 rad/s; the end moves at 6.0 m/s.
4. Common mistakes
A hoop and a disk have the same mass and radius and spin at the same ω about their centers. Which has more kinetic energy?
K = ½Iω². The hoop’s I (MR²) is twice the disk’s (½MR²), so its kinetic energy is twice as large.
Ready to practice?
4 questions on Rotational Energy & Work, each with a hint and an explanation for every wrong answer.