A rolling body carries both translational and rotational kinetic energy, so its total is KE = 0.5 * m * v^2 + 0.5 * I * omega^2, linked by the rolling condition v = omega * R. The arithmetic is exact only for the stated mass, speed, inertia, and angular speed in the ideal rigid-body, no-slip model. Real systems add distributed-mass geometry, slipping, friction, bearing losses, axis alignment, sensor calibration, finite precision, and measurement uncertainty.

A rolling object carries two checked kinetic-energy ledgers. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Rolling energy has translation plus rotation

The translational part is 9 joules and the rotational part is 9 joules.

9+9=189+9=18
Rolling energyTranslation and rotation are separate ledgers.r=1 mm=2 kgr=1 mI=2 kg*m^2omega=3 1/sKrot=9 JKtrans=9 JKtotal=18 J

The total is the sum of the two ledgers

The rolling total is 18 joules.

Ktotal=18 JK_{\mathrm{total}}=18\ \mathrm{J}
Rolling energyTranslation and rotation are separate ledgers.r=1 mm=2 kgr=1 mI=2 kg*m^2omega=3 1/sKrot=9 JKtrans=9 JKtotal=18 J