A solid sphere and a solid cylinder of the same mass and radius roll without slipping. If their radii of gyration satisfy kₛₚₕ:k꜀ᵧₗ = 2:√x, find x.

Question

G1 Physics · Accepted Answer

For a solid sphere: I = (2/5)MR² = Mkₛₚₕ² So: kₛₚₕ = R√(2/5) For a solid cylinder about its symmetry axis: I = (1/2)MR² = Mk꜀ᵧₗ² So: k꜀ᵧₗ = R/√2 Therefore: kₛₚₕ/k꜀ᵧₗ = √(2/5) × √2 = 2/√5 Given: kₛₚₕ:k꜀ᵧₗ = 2:√x Hence: √x = √5 x = 5 Final answer: 5.