Rolling motion

An object rolling without slipping is doing two things at once: its centre of mass translates forward, and the object spins about that centre, and the "no slipping" condition locks the two together — the point touching the ground is momentarily at rest, which forces v = ωr exactly. Because both motions happen simultaneously and neither can be dialled independently, the object's total kinetic energy has two terms, not one: ½mv² for the centre of mass moving along, plus ½Iω² for the spin about that centre. A sliding block only ever has the first term; a rolling object always has both.

That second term is where energy hides, and how much it takes depends on how the object's mass is distributed — its moment of inertia. A solid ball has most of its mass close to the rotation axis and a comparatively small moment of inertia; a hollow shell of the same mass and radius has its mass pushed out to the rim and a much larger one. Released from rest at the same height, both convert the same total gravitational potential energy, but the shell diverts a larger fraction of it into spinning and less into moving forward, so it always loses the race — not because of friction robbing it of energy, but because more of its energy went somewhere other than translational motion.

That same accounting is why any rolling object is slower at the bottom of a slope than a frictionless block sliding the identical drop: the block converts all its potential energy into translational kinetic energy, while the rolling object must divide the same potential energy between translation and spin. Rolling friction here does no work and dissipates no energy at all — it is what enforces v = ωr rather than allowing slipping, not a drag force stealing energy from the motion.