Use 1 + k²/R² for Every Rolling Problem: Speed, Acceleration, Time and Energy

The total kinetic energy of a body rolling without slipping is (1/2) m v² (1 + k²/R²). The fraction that is rotational is (k²/R²)/(1 + k²/R²): for a ring 1/2, for a disc 1/3, for a solid sphere 2/7. So a body with 50% of its energy in rotation is a ring. Rotational energy (1/2) I ω² scales with I when ω is equal: for a solid sphere, a disc and a ring of the same mass, radius and angular speed, the work to stop them is greatest for the ring, then the disc, then the sphere. A sphere and a disc with equal radii and masses, the disc turning at twice the sphere's angular speed, have kinetic energies in the ratio 1 : 5.

Therefore g > a_sphere > a_disc > a_ring, and the sphere reaches the bottom first whatever the masses are. The time ratio of a sphere to a disc is sqrt(7/5) : sqrt(3/2) = sqrt(14) : sqrt(15).

In pure rolling, the centre moves at v and the point at the top moves at 2v, the point of contact is at rest, and the points at the ends of the horizontal diameter move at v sqrt 2, so for v = 5 m/s they move at 5 sqrt(2) m/s. A solid cylinder of mass 12 kg rolling at 4 m/s into a spring of constant 200 N/m compresses it until all its kinetic energy (1/2) m v² (1.5) = 144 J becomes (1/2) k x², so x = 1.2 m.