Rotational Kinetic Energy & Work

You’ll be able to

Compute rotational kinetic energy as KE_rot = ½Iω²
Find the total kinetic energy of a rolling object as translational plus rotational
Apply the rotational work-energy relationship W = ∫τ dθ

Spinning stores kinetic energy

A rotating body carries kinetic energy even if its center of mass is still. Summing ½(dm)v² over the body, with v = rω, gives KE_rot = ½Iω² — the exact analog of ½mv², with the moment of inertia in place of mass and angular speed in place of linear speed. A rolling object has both: its total kinetic energy is ½mv_cm² (translation) plus ½Iω² (rotation).

Rotational and total kinetic energy

KE_rot = ½Iω² KE_total = ½mv_cm² + ½Iω²

The rotational twin of ½mv². For rolling, add the translational and rotational parts, linked by v_cm = Rω.

Work and power in rotation

A torque acting through an angular displacement does work: W = ∫τ dθ, the rotational version of ∫F dx. This work changes the rotational kinetic energy, giving a rotational work-energy theorem W_net = ΔKE_rot. The rate at which a torque delivers energy is the rotational power P = τω, the twin of P = Fv.

Rotational work and power

W = ∫ τ dθ P = τω

Torque integrated over angle gives work; torque times angular velocity gives power. Both mirror their translational forms exactly.

Worked example

A uniform solid disk (I = ½MR²) of mass 2 kg rolls without slipping at v_cm = 3 m/s. Find its total kinetic energy.

1.Translational part: ½mv_cm² = ½(2)(3²) = 9 J.
2.Rotational part: ½Iω². With ω = v/R and I = ½MR², this is ½(½MR²)(v/R)² = ¼Mv² = ¼(2)(9) = 4.5 J.
3.Add the two contributions: KE_total = 9 + 4.5 = 13.5 J.
4.Notice the rotational share is exactly one third of the total for a disk.

Answer: KE_total = ½mv_cm² + ½Iω² = 9 + 4.5 = 13.5 J

Checkpoint

A flywheel has a moment of inertia of 4 kg·m² and spins at 3 rad/s. What is its rotational kinetic energy?

Checkpoint

A solid disk (I = ½MR²) rolls without slipping. What fraction of its total kinetic energy is rotational?

Tip

When a rolling object descends, set mgh = ½mv_cm² + ½Iω² and substitute ω = v/R. The rotational term makes rolling objects reach the bottom slower than a frictionless sliding block, because some energy is tied up in spin.

Answer the 2 checkpoints as you read.