A solid cylinder of mass 0.750 kg and radius 4.00 cm rolls from rest without slipping down an incline 2.00 m high. What is its speed at the bottom?

1. Conservation of energy gives m g h = one half m v squared plus one half I omega squared. 2. For a solid cylinder, I = one half m R squared. 3. Rolling without slipping gives omega = v / R, so the rotational term becomes one quarter m v squared. 4. Substituting, m g h = one half m v squared plus one quarter m v squared = three quarters m v squared. 5. Mass and radius both cancel, leaving v = square root of (4 g h / 3). 6. Substituting, v = square root of [4(9.80 m/s^2)(2.00 m) / 3] = square root of 26.13 m^2/s^2. 7. So v = 5.11 m/s, and the sliding answer square root of 2 g h would have been 6.26 m/s. _Source: OpenStax College Physics (CC BY 4.0), Ch 10 "Rotational Motion and Angular Momentum", section 10.4 Rotational Kinetic Energy: Work and Energy Revisited_