A solid sphere and a thin hoop, each of mass 0.7 kg, roll from rest down the same slope through a vertical drop 0.4 m. They roll without slipping; ignore air resistance and rolling losses. For each object v = ωR; I_sphere = 2mR²/5 and I_hoop = mR². Use g = 9.81 m s⁻². Find each final speed and the fraction of each object's kinetic energy that is rotational. Explain which is faster.

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Sphere 2.368 m s⁻¹, rotational fraction 2/7; hoop 1.981 m s⁻¹, rotational fraction 1/2. The sphere is faster. Mechanical energy is conserved because the stated model excludes dissipative losses. Static friction enforces rolling but the contact point is instantaneously at rest, so it does not dissipate mechanical energy. Write mgh = ½mv² + ½Iω² and substitute I = βmR² and ω = v/R. This gives v² = 2gh/(1 + β); mass and radius cancel. For the sphere β = 2/5, so v = √(10gh/7) = 2.368 m s⁻¹. For the hoop β = 1, so v = √(gh) = 1.981 m s⁻¹. The rotational fraction is β/(1 + β), giving 2/7 for the sphere and 1/2 for the hoop. A larger fraction assigned to rotation leaves less translational kinetic energy for the same drop. The sphere therefore reaches the larger speed. Both speeds are below √(2gh), the speed of a frictionless point mass, which is a useful energy check.

A sphere and a hoop roll downhill · set 2

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